LIBRARY 

OF  THE 

UNIVERSITY  OF  CALIFORNIA. 


Class 


A  TEXT-BOOK 

OP 

ENGINEERING 


A  TEXT-BOOK 


OF 


ENGINEERING 


FOR 


SECONDARY 
TECHNICAL  SCHOOLS 


BY 


WILLIAM  R.  KING,  U.  S.  N.,  Retired 

Principal  and  Head  of  Department  of  Engineering 
Baltimore  Polytechnic  Institute 


PUBLISHED  BY 
THE  FRIEDENWALD  COMPANY 

BALTIMORE,   MD. 
1906 


COIERAL 


COPYRIGHT,  1906 

BY 
WILLIAM  R.  KING 


jfrufcennrnffc 

BALTIMORE,  MD.,  U.  S.  A. 


PREFACE 

The  preparation  of  this  text-book  is  the  result  of  four  years' 
experience  in  the  class-room  in  giving  students  of  a  secondary 
technical  school  of  the  highest  grade  the  maximum  of  engineering 
instruction  admissible,  considering  the  time  at  disposal  and  the 
maturity  of  the  students. 

It  is  believed  that  the  simplicity  employed  in  the  presentation  of 
the  elements  of  the  different  subjects  gives  an  excellent  ground- 
work upon  which  to  build,  should  the  graduate  fail  to  pursue  to 
completion  his  technical  education  at  a  university.  In  the  event 
of  his  entry  into  a  university,  it  is  established  by  precedent  that  he 
should  be  able  to  obtain  a  degree  in  two  years. 

It  is  intended  that  the  first  six  chapters  of  Part  L,  supplemented 
by  some  elementary  text  on  the  constructive  details  of  steam 
machinery — such  as  that  by  Spangler,  Greene,  and  Marshall — shall 
constitute  the  course  for  the  third  year  students;  the  remainder  of 
the  book  is  to  be  covered  during  the  fourth  year. 

Correlative  with  the  subject-matter  of  the  book,  it  will  be  found 
that  there  is  necessity  for  a  course  in  mathematics,  including  the 
Calculus,  and  one  in  Demonstrative  Mechanics,  together  with  facili- 
ties for  experimental  work  in  a  fairly  equipped  mechanical 
laboratory. 

The  article  in  the  Appendix  on  the  distribution  of  steam  in  a 
cylinder  and  the  formation  of  the  indicator  diagram  was  contributed 
by  Mr.  W.  L.  DeBaufre  of  the  class  of  1903,  Baltimore  Polytechnic 
Institute,  now  a  senior  at  Lehigh  University. 

Due  acknowledgment  is  made  to  Mr.  S.  P.  Platt,  Instructor  in 
charge  of  Mechanical  Drawing  at  the  Baltimore  Polytechnic  Insti- 
tute, for  assistance  in  making  the  drawings. 


iv  PREFACE 

Much  of  the  matter  on  Iron  and  Steel  in  Chapter  I.,  Part  III., 
has  been  taken  from  Durand's  •Practical  Marine  Engineering,  by 
permission  of  the  author. 

Works  on  engineering  subjects  have  been  consulted,  including 
those  of  Sennett,  Seaton,  Holmes,  Merriman,  Goodman,  and 
Jamieson. 

WILLIAM  E.  KING, 
'Passed  Assistant  Engineer,  U.  S.  N.f  Retired. 

BALTIMORE  POLYTECHNIC  INSTITUTE, 
January,  1906. 


CONTENTS 

PART  I 

I.     INTRODUCTORY  

II.     THE  APPLICATION  OF  HEAT  TO  WATER n 

III.     FUELS  AND  COMBUSTION 


ERRATA 

A  few  errors  escaped  notice  while  passing  through  the  press.  On 
page  3,  last  line  of  Art.  5,  read  on  the  Fahrenheit  scale,  the  tem- 
perature of  the  original  volume  heing  that  of  melting  ice.  On 
page  141,  second  line  of  last  paragraph,  read  AB  instead  of  AD. 
On  page  146,  last  line,  read  stroke  of  the  piston  instead  of  travel 
of  the  valve.  On  page  170,  fifth  line  from  the  bottom,  read  that 
instead  of  than.  On  page  189,  sixth  line  from  the  top,  read  steam 
instead  of  stream.  On  page  288,  third  line  from  bottom  read  W3 
instead  of  W.  On  page  290,  in  the  value  of  Mw?,  read  Ws  (dj  +  d2) 
instead  of  W^  X  d2.  On  page  322,  fourth  line  from  bottom,  read 
K2  instead  of  I.  On  page  325,  fifth  line  from  bottom,  read  7,OOOD:5 
instead  of  7,OOOD.  In  Example  V,  page  326,  the  factor  10,000 
has  been  omitted  under  the  radical.  On  page  332,  in  the  value  of 
Ely,  read  Lx3  instead  of  wx3.  On  page  334,  in  the  value  of  M, 
read  w  instead  of  W.  On  page  338,  third  line  from  bottom,  read 
SL2  instead  of  SL3.  In  Art.  36,  page  366,  read  2n  —  2  instead  of 
n  —  2.  On  page  371,  seventh  line  from  bottom,  read  a  instead  of  a. 


VII.    RESILIENCE    ...  **» 

337 

PART  V 

I.    Bow's  SYSTEM  OF  LETTERING.     FORCE  DIAGRAM.     FUNICULAR 
.POLYGON 

II.     FRAMED  STRUCTURES.    RECIPROCAL  DIGRAM  " 
APPENDIX  

INDEX  38^ 

• 395 


iv  PREFACE 

Much  of  the  matter  on  Iron  and  Steel  in  Chapter  I.,  Part  III., 
has  been  taken  from  Durand's  •Practical  Marine  Engineering,  by 
permission  of  the  author. 

Works  on  engineering  subjects  have  been  consulted,  including 
those  of  Sennett,  Seaton,  Holmes,  Merriman,  Goodman,  and 
Jamieson. 


CONTENTS 

CHAP.  PART    I  PAGE. 

I.  INTRODUCTORY  3 

II.     THE  APPLICATION  OF  HEAT  TO  WATER 11 

III.  FUELS  AND  COMBUSTION 22 

IV.  EFFICIENCY   33 

V.     THE  VALVE  AND  ITS  MOTION 45 

VI.     THE   CONVERSION   OF  MOTION.     ACTION   OF   THE   CRANK   AND 

CONNECTING-ROD 64 

VII.     STAGE  EXPANSION  ENGINES 70 

VIII.     THE  INDICATOR  AND  ITS  DIAGRAM 75 

IX.  MEAN  PRESSURE  OF  A  GAS  EXPANDING  WITHIN  A  CYLINDER.  ...  96 

X.     BOILER  AND  ENGINE  EFFICIENCY 103 

XL     ENGINE  DESIGN   121 

XII.     THE  ZEUNER  VALVE  DIAGRAM 137 

XIII.  ECONOMY  OF  THE  STEAM  ENGINE  AND  BOILER 150 

XIV.  STEAM  BOILERS   179 

XV.     THE  STEAM  TURBINE 190 

PART  II 

I.     BELTING   205 

II.     WHEELS  IN  TRAIN 217 

PART  III 

I.    MATERIALS 231 

II.     TESTING  MATERIALS 255 

PART  IV 

I.  MOMENTS.     CENTER  OF  GRAVITY 267 

II.  BENDING  MOMENT.   SHEAR.   BENDING-MOMENT  DIAGRAM.   SHEAR 

DIAGRAM 282 

III.  MOMENT  OF  INERTIA.    RADIUS  OF  GYRATION 304 

IV.  THE  THEORY  OF  BEAMS 311 

V.  COLUMNS.    SHAFTS 320 

VI.  THE  DEFLECTION  OF  BEAMS 329 

VII.  RESILIENCE    337 

PART  V 

I.     Bow's  SYSTEM  OF  LETTERING.      FORCE  DIAGRAM.     FUNICULAR 

POLYGON 345 

II.  FRAMED  STRUCTURES.    RECIPROCAL  DIAGRAM 362 

APPENDIX  383 

INDEX     .   395 


PART  I 

THE   ELEMENTS  OF  STEAM 
ENGINEERING 


JNIVERSITY 


CHAPTEE  I. 
INTRODUCTORY. 

1.  A  gas  may  be  defined  as  a  fluid  of  such  nature  that  if  a  certain 
volume  of  it  be  admitted  into  a  vessel,  the  volume  admitted  will 
distribute  itself  throughout  the  vessel,  whatever  the  volume  of  the 
vessel  may  be. 

2.  The  Kinetic  Theory  of  Gases  is  that  the  molecules  of  a  gas 
are  in  a  state  of  rapid  motion  in  straight  lines,  and  that  as  a  result 
of  their  collision  with  each  other  and  with  the  walls  of  the  contain- 
ing vessel,  pressure  is  exerted. 

3.  The  two  laws  of  gaseous  expansion  which  are  of  wide  applica- 
tion are  those  of  Boyle  and  of  Charles. 

4.  Boyle's  Law.  —  The  law  of  Boyle  may  be  stated  as  follows  : 
The  pressure  of  a  mass  of  gas  at  constant  temperature  varies 

inversely  as  its  volume. 

This  law  may  also  be  expressed  as  follows  : 

The  product  of  the  pressure  and  volume  of  a  mass  of  gas  is  a 
constant  quantity  as  long  as  the  temperature  is  constant. 
The  algebraic  expression  for  Boyle's  law  is 

PV  =  C 

in  which  P  is  the  pressure  in  any  units,  usually  in  pounds  per  sq. 
foot,  V  the  volume  in  cubic  feet,  and  C  a  constant  quantity  to  be 
determined  experimentally  for  a  particular  gas. 

5.  Charles's  Law,  known  also  as  the  law  of  Gay  Lussac.  —  This 
law  asserts  that  all  gases  have  the  same  coefficient  of  expansion,  and 
this  coefficient  is  the  same  whatever  the  pressure  supported  by  the 
gas.     Hence  for  each  degree  of  rise  or  fall  of  temperature  of  a  gas 
its  volume  will  be  increased  or  diminished  by  a  fixed  fraction  of 
its   original    volume.     This    fraction    has    been    computed    to    be 

2^  —  0.0036625  on  the  Centigrade  scale,  and  Tcrr-g  =  0.0020347 
on  the  Fahrenheit  scale. 

6.  As  in  the  case  of  the  air  thermometer,  if  a  column  of  air  be 
inclosed  in  a  tube  of  uniform  bore,  and  separated  from  the  atmos- 


4  THE.  ELEMENTS  OF  STEAM  ENGINEERING 

phere  by  an  air-tight  piston  which  is  free  to  move  in  the  tube ;  and 
if  when  the  temperature  of  the  confined  air  is  that  of  melting  ice 
the  height  of  the  column  is  unity,  then,  if  the  tube  be  exposed  to 
the  steam  of  boiling  water,  the  pressure  remaining  constant,  the 
temperature  of  the  inclosed  air  will  rise  to  100°  C.,  or  to  212°  F. 
The  volume  of  the  air  will  then  be,  by  Charles's  law, 

1  +  (100  X  .0036625)  =  1.36625 
or  1  +  (212  —  32) .0020347  =  1.36625  . 

This  unit  volume  of  air  having  expanded  the  fraction  0.36625  of 
itself  by  raising  its  temperature  100°  C.,  or  180°  F.,  it  follows  that 
its  volume  will  be  doubled  when  the  temperature  is  raised  through 
x° ,  thus: 

0.36625  :  1  :  :  100  :  x,  whence  x  =  273°  C. 
or  0.36625  :  1  :  :  180  :  x,  whence  x  =  491.6°  F. 

It  follows,  by  the  law  of  Charles,  that  if  the  unit  volume  had  been 
cooled  through  273°  C.,  or  through  491.6°  F.,  there  would  be  no 
existing  volume. 

This  point  is  called  the  absolute  zero  of  temperature,  and  tem- 
peratures reckoned  from  this  point  are  called  absolute  tempera- 
tures. 

The  absolute  zero  is  then  273°  below  the  zero  of  the  Centigrade 
scale,  and  491.6  —  32  =  459.6°  below  the  zero  of  the  Fahrenheit 
scale.  To  convert  temperatures  measured  on  the  C.  or  F.  scales  to 
absolute  temperatures,  we  have  only  to  add  273°,  or  459.6°,  as  the 
case  may  be.  Absolute  temperatures  will  be  denoted  by  the  capital 
letter  T. 

It  must  be  observed  that  if  the  volume  of  a  gas  be  kept  constant 
and  its  temperature  raised,  the  pressure  will  be  increased  or  dimin- 
ished in  the  same  ratio  that  the  volume  was  increased  or  diminished 
when  the  pressure  was  constant.  For  let  P  and  V  be  the  pressure 
and  volume  of  a  portion  of  gas  at  32°  F.,  and  1.366257  the  volume 
when  heated  to  212°  F.  under  the  constant  pressure  P.  If  the 
temperature  of  the  gas  be  now  kept  constant  at  212°  while  it  is 
being  compressed  back  to  its  original  volume  7,  it  will  then  have 
a  certain  pressure  P',  and  we  will  have,  by  Boyle's  law, 

P'V  =  P  X  1.366257,  whence  P'  —  1.36625P. 

In  other  words,  if  a  volume  7  of  a  gas  at  32°  and  pressure  P  be 
heated  to  212°,  the  volume  remaining  at  7,  its  pressure  will  be 
1.36625P. 


INTRODUCTORY  5 

We  may  now  regard  this  absolute  zero  from  another  viewpoint. 
Let  us  consider  a  mass  of  perfect  gas  of  volume  v0,  pressure  p0,  and 
a  temperature  0°  C.,  or  32°  F.  Imagine  the  volume  kept  constant 
while  its  temperature  is  raised  or  lowered  t°,  the  resulting  pressure 
p  will  be,  by  Charles's  law,  p  =  p0  ±  p0at  =  p0(l  ±  at},  where  a 
is  the  coefficient  of  expansion.  If  the  gas  now  be  cooled  until  its 

1° 
temperature  is  reduced  to  -  -  — ,  we  will  then  have : 


1° 
That  is,  at  a  temperature  -      -  the  gas  would  exert  no  pressure 

on  the  walls  of  the  containing  vessel.  If  the  kinetic  theory  of  gases, 
or  the  theory  that  heat  is  a  form  of  energy,  or  of  motion,  be 
accepted,  then  when  the  point  is  reached  where  molecular  motion 
ceases,  all  heat  and  pressure  must  be  non-existent;  and  as  it  is 
impossible  to  imagine  a  body  colder  than  one  devoid  of  heat,  that 

1° 
is,  one  at  a  temperature  —  —  ,  this  temperature  is  called  the  abso- 

lute zero. 

7.  Combination  of  the  Laws  of  Boyle  and  Charles.  —  A  very  use- 
ful equation  in  the  solution  of  questions  concerning  the  pressure, 
volume,  and  temperature  of  gases  may  be  obtained  from  a  com- 
bination of  the  laws  of  Boyle  and  Charles. 

Let  v0  be  the  volume  and  p0  the  pressure  of  a  portion  of  gas  at 
0°  temperature.  Let  v  -denote  its  volume  when  the  temperature 
is  raised  to  if,  the  pressure  remaining  at  p0,  and  let  v'  be  the  volume 
under  the  pressure  p',  the  temperature  remaining  if.  Then  by 
Boyle's  law  we  shall  have  p'v'  =  p0v  =  C.  By  Charles's  law  we 
have  v  =  v0(l  +  at').  Hence  : 


But  —  =  T0  =  the  absolute  zero  of  temperature  ;    hence  a  =  -^  , 
and  1  +  f  =  T  ;  therefore  p'v'  =  (£&}  T<  ,  or  #  = 


In  other  words,  the  product  of  the  volume  and  pressure  of  a  gas  is 
proportional  to  the  absolute  temperature. 

By  Boyle's  law  we  have,  PV  =  a  constant  in  foot-pounds,  where 
P  =  pressure  in  pounds  per  square  foot,  and  V  —  volume  in  cubic 


6  THE  ELEMENTS  OF  STEAM  ENGINEERING 

feet.     The  constants  for  different  gases  have  been  calculated  by 
Kegnault  with  great  accuracy. 

The  constant  for  air  may  be  found  as  follows  :  It  is  known  that 
water  is  773  times  as  heavy  as  air,  the  temperature  of  which  is 
32°  F.,  or  0°  C.,  and  the  pressure  14.7  Ibs.,  or  760  mm.;  and  since 

773 
a  cubic  foot  of  water  weighs  62.4  Ibs.,  it  will  take  —  12-387 


cubic  feet  of  air  to  weigh  one  pound.  This  pound  of  air  may  be 
assumed  to  be  contained  in  a  c}dinder  1  sq.  ft.  in  area  and  12.387 
feet  in  height.  The  pressure  of  the  air  being  14.7  Ibs.  per  sq.  inch, 
we  may  assume  at  the  bottom  of  the  cylinder  a  piston  weighing 
144  X  14.7  pounds  —  P.  The  movement  of  this  piston  through 
the  height  of  the  cylinder  will  displace  12.387  cu.  ft.  =  V,  and 
perform  144  X  14.7  X  12.387  foot-pounds  of  work.  Hence,  PV  = 
Constant  —  144  X  14.7  X  12.387  =  26,220  ft.  Ibs. 

We  have  shown  that  the  pressure  multiplied  by  the  volume  of  a 
portion  of  gas  is  proportional  to  the  absolute  temperature;  so  that 
if  T0  be  the  absolute  zero  on  the  Fahrenheit  scale  (491.6°  below 
32°)  corresponding  to  P0V0,  then  for  any  other  pressure  and  vol- 
xume,  as  P  and  V,  we  shall  have  : 

-> 

wlience  FT'  =  53.3T'. 

This  equation  for  a  perfect  gas  is  usually  written  PV  =  RT  , 
where  R  is  a  constant  depending  on  the  density  of  the  gas.  For 
air  we  have  found  it  to  be  53.3.  For  superheated  steam  it  is  85.5. 

8.  The  Caloric  or  Material  Theory  of  Heat  was  advanced  by  Black 
in  1798,  but  in  1802  it  wTas  shown  that  heat  could  be  produced  to  an 
unlimited  extent  by  friction,  thus  proving  its  identity  with  motion. 

9.  The  Kinetic  Theory  of  Heat  teaches  that  heat  is  a  form  of 
motion  consisting  of  the  agitation  of  the  molecules  of  matter,  either 
iby  rotation  or  by  vibration,  and  the  hottest  bodies  are  those  whose 
molecules  are  agitated  at  the  greatest  velocities. 

Another  statement  of  the  kinetic  theory  is  that  heat  is  a  form 
of  energy,  or  that  it  is  energy  in  the  invisible  form  of  molecular 
motion. 

Since  the  motion  of  the  molecules  of  a  body  is  invisible,  the  exact 
nature  of  molecular  motion  is  unknown,  but  the  application  of  the 
laws  of  motion  and  energy  to  phenomena  produced  by  heat  has 


INTRODUCTORY  7 

produced  the  knowledge  upon  which  is  founded  the  science  of 
Thermodynamics,  or  the  Mechanical  Theory  of  Heat. 

10.  Energy  is  defined  as  the  capacity  to  do  work.     A  force  acting 
through  a  distance  exerts  energy,  and  the  resistance  overcome  is  the 
work  done. 

Energy  is  of  two  kinds — Kinetic  and  Potential.  The  energy 
of  a  body  due  to  its  motion  is  kinetic,  and  is  measured  by  the  work 
done  by  the  body  in  being  brought  to  rest.  The  work  done  in 
raising  a  body  of  weight  w  through  a  height  h  is  ivh,  and  if  v  be 
the  velocity  acquired  by  the  body  in  falling  a  distance  In,  then 

7          wvz        mv2       i  •  i    •     ,1 
v2  =  2gh,  whence  ivli  -     -3 —  =  —w~,  which  is  the  expression  for 

the  kinetic,  or  actual,  energy  of  a  body  of  weight  w,  moving  at  a 
velocity  v.  It  is  in  fact  the  work  stored  up  in  a  body  of  weight  w, 
moving  with  a  velocity  v. 

If  a  body  be  raised  a  distance  above  the  surface  of  the  earth  a 
certain  amount  of  work  is  performed  in  overcoming  the  attraction 
of  gravity,  and  this  work,  stored  in  the  body,  gives  to  it  the  power 
of  doing  work.  Energy  thus  possessed  by  a  body,  due  to  its  posi- 
tion and  not  its  velocity,  is  called  potential  energy.  A  body  may 
possess  potential  energy  due  to  the  action  of  forces  other  than 
gravity,  as,  for  example,  in  winding  a  watch,  work  is  done  on  the 
spring  by  which  its  energy  is  increased,  giving  to  it  potential 
energy,  which  causes  a.reappearance  of  the  work  in  driving  the  watch 
while  the  spring  unbends. 

11.  Temperature. — Temperature  refers  to  the  condition  of  a  body 
as  regards  its  sensible  heat,  and  this  condition  is  measured  by  the 
thermometer.     Two  bodies  are  said  to  be  of  equal  temperature  when 
there  is  no  tendency  of  heat  to  pass  from  one  to  the  other. 

12.  Thermometer. — A  thermometer  is  an  instrument  for  measur- 
ing temperatures.     The  lower  fixed  point  on  thermometric  scales  is 
the  temperature  of  melting  ice  under  atmospheric  pressure.     The 
upper  fixed  point  is  the  temperature  of  boiling  water  under  the  nor- 
mal pressure  of  the  atmosphere.     The  scale  universally  used  for 
scientific  work  is  that  known  as  Centigrade,  but  in  England  and  in 
the  United  States  the  Fahrenheit  scale  is  in  general  use.     In  Eussia 
the  Eeaumur  scale  is  used. 

The  temperature  of  melting  ice  is  marked  0°  on  the  Centigrade 
and  Eeaumur  scales,  but  on  the  Fahrenheit  scale  it  is  marked  at  32  °t 


8  THE  ELEMENTS  OF  STEAM  ENGINEERING 

The  temperature  of  boiling  water  is  marked  80°  on  the  Keaumur, 
100°  on  the  Centigrade,  and  212°  on  the  Fahrenheit  scales.  Denot- 
ing the  three  scales  by  (7,  F,  and  R,  we  shall  have  0°  C  =  32°  F  = 
0°  R  .  The  zero  of  the  Fahrenheit  scale  is  then  32°  below  the  tem- 
perature of  melting  ice,  so  if  32°  be  deducted  from  the  Fahrenheit 
reading  we  shall  get  the  number  of  degrees  on  each  scale  by  which 
any  given  temperature  differs  from  that  of  melting  ice.  Let  the 
given  temperature  be  that  of  boiling  water  :  then 

C  :  F  —  32  :  R  :  :  100  :  180  :  80  :  :  5  :  9  :  4 


5  9         ~  4  * 

By  means  of  these  equations  temperatures  on  one  scale  may  readily 
be  converted  into  either  of  the  others. 

13.  Calorimetry.  —  In   order   to   measure   the   quantity   of  heat 
which  a  body  absorbs  or  yields  in  passing  through  a  given  range  of 
temperature,  or  in  changing  its  state,  we  adopt  as  a  unit  that  quan- 
tity of  heat  which,  acting  on  a  given  weight  of  water,  changes  its 
temperature  a  definite  amount.    The  apparatus  in  which  the  meas- 
urement is  made  is  called  a  calorimeter.     The  specific  heats,  the 
latent  heats  of  vaporization,  and  the  heats  of  combustion  of  sub- 
stances are  determined  by  the  calorimeter. 

14.  Thermal  Unit.—  The  unit  of  Heat,  or  the  British  Thermal 
Unit,  is  the  quantity  of  heat  required  to  raise  the  temperature  of 
one  pound  of  water  through  one  degree  Fahrenheit,  when  at  or  near 
its  temperature  of  greatest  density,  39°.     More  recently  the  meas- 
ure of  the  unit  is  taken  when  the  water  is  at  a  temperature  of  62°. 

15.  Specific  Heat.  —  The  specific  heat  of  a  substance,  or  its  capac- 
ity for  heat,  is  the  ratio  of  the  quantity  of  heat  required  to  raise 
the  temperature  of  a  given  weight  of  the  substance  through  1°  F., 
to  the  quantity  required  to  raise  the  temperature  of  an  equal  weight 
of  water  at  39°  F.  through  1°  F. 

16.  Thermodynamics.  —  The  science  which  treats  of  the  mechani- 
cal or  dynamical  theory  of  heat,  or  of  the  relation  between  heat  and 
work,  is  called  thermodynamics. 

17.  First  Law  of  Thermodynamics.  —  Heat  and  mechanical  energy 
are  mutually  convertible,  and  heat  requires  for  its  production,  or 
produces  by  its  disappearance,  mechanical  energy  equivalent  to  778 
foot-pounds  for  each  thermal  unit. 


INTRODUCTORY  9 

18.  Mechanical  Equivalent  of  Heat. — One  thermal  unit  is  equiv- 
alent to  778  foot-pounds  of  mechanical  work. 

19.  The  Conservation  of  Energy. — The  first  law  of  thermody- 
namics expresses  the  principle  of  the  conservation  of  energy — that 
the  total  energy  of  the  universe  is  constant,  that  energy  can  neither 
be  created  nor  destroyed  by  any  process  of  nature,  and  that  every 
gain  in  one  form  of  energy  corresponds  exactly  to  a  loss  in  some 
other  form.     While  no  heat  is  absolutely  lost  in  the  transformation 
of  energy  of  one  form  into  that  of  another,  a  considerable  amount  is 
lost  as  available  energy,  being  transformed  into  useless  heat  and 
dissipated,  or  diffused  among  surrounding  bodies  and  thus  becoming 
unavailable.     It  is  only  when  a  body  is  hotter  than  surrounding 
objects  that  its  heat  energy  can  be  utilized  in  doing  work.     If  all 
bodies  were  of  the  same  temperature  no  work  of  any  kind  would  be 
possible. 

A  general  statement  of  the  conservation  of  energy  is : 

Kinetic  Energy  +  Potential  Energy  —  Constant, 
but  the  practical  statement  is :   In  any  machine  we  have :   The  total 
energy  deposited  =  Useful  work  given  out  -f-  Work  lost  by  resist- 
ances.    This   disposes   of  the  possibility   of   "perpetual   motion," 
since  a  machine  cannot  deliver  more  work  than  was  deposited. 

20.  Second  Law  of  Thermodynamics. — Heat  cannot  pass  from  a 
cold  body  to  a  hot  one  by  a  self-acting  process  unaided  by  external 
agency. 

It  follows  from  this  law  that  no  engine  can  convert  into  work 
the  whole  of  the  heat  supplied  to  it;  for  as  soon  as  the  heat  sup- 
plied has  fallen  in  temperature  to  that  of  the  surrounding  atmos- 
phere the  heat  remaining  is  no  longer  available  for  doing  useful 
work.  It  follows  also  from  this  law  that  it  is  impossible  to  convert 
any  part  of  the  heat  of  a  body  into  mechanical  work,  except  by 
allowing  it  to  pass  to  another  body  at  a  lower  temperature. 

21.  Definition. — Force  is  that  which  moves  or  tends  to  move  a 
body,  or  which  changes  or  tends  to  change  the  motion  of  a  body. 

The  unit  of  a  force  is  a  pound,  and  forces  are  expressed  as  being 
equal  to  so  many  pounds. 

22.  Definition. — Work  is  the  overcoming  of  resistance  through 
space,  and  its  measure  is  the  product  of  the  resistance  and  the  space 
through  which  it  is  overcome,  the  space  being  usually  reckoned  in 
feet. 


10  THE  ELEMENTS  OF  STEAM  ENGINEERING 

23.  Unit  of  Work. — Since  the  unit  of  force  is  a  pound,  and  the 
unit  of  distance  a  foot,  the  product  formed  by  multiplying  a  foot 
by  a  pound  is  the  unit  of  work,  and  is  called  a  foot-pound.     A  foot- 
pound represents  the  work  done  in  raising  the  weight  of  a  pound 
through  the  distance  of  one  foot.     If  10  pounds  be  raised  5  feet, 
or  5  pounds  be  raised  10  feet,  the  work  done  in  either  case  is  rep- 
resented by  50  foot-pounds. 

Other  units  of  work  are  sometimes  used  as  a  matter  of  con- 
venience. For  example,  an  inch-ton  represents  the  work  performed 
in  raising  2240  pounds  one  inch,  and  it  is  equal  to  2240  inch- 
pounds,  or  to  ..Q  foot-pounds.  A  foot-ton  represents  an  expendi- 
ture of  work  equal  to  2240  foot-pounds.  It  is  thus  seen  that  the 
different  units  of  work  are  readily  converted  one  to  another. 

24.  Definition. — Power  is  the  rate  of  doing  work,  and  is  meas- 
ured by  the  number  of  foot-pounds  of  work  done  in  a  unit  of  time. 

25.  Definition. — Horse-power  is  the  unit  employed  to  represent 
the  rate  of  work  of  a  steam  engine,  and  its  measure  is  the  quantity 
of  work  equivalent  to  the  raising  of  33,000  pounds  through  one  foot 
in  one  minute.     The  performance  of  33,000  foot-pounds  of  work 
in  one  minute  is  then  a  horse-power,  and  an  engine  performing  this 
amount  of  work  in  a  second  wrould  be  working  at  the  rate  of  60 
horse-power. 

26.  Definition. — Brake  Horse-power  is  the  power  which  the  en- 
gine transmits  for  useful  work ;  that  is,  it  is  the  total,  or  Indicated 
Horse-power,  minus  the  powrer  absorbed  in  driving  the  engine  itself. 

PEOBLEMS. 

1.  The  ratio  of  the  volumes  of  air  under  the  same  pressure  at 
temperatures  of  0°  C.  and  of  100°  C.,  or  at  32°  F.  and  at  212°  F., 
is  1  to  1.36625.     Find  the  absolute  zero  of  temperature  on  the 
Centigrade  and  Fahrenheit  scales. 

2.  A  volume  of  air  at  temperature  of  32°  F.,  and  pressure  of  14.7 
Ibs.  per  sq.  inch,  is  heated  to  a  temperature  of  100°  F.,  the  volume 
remaining  the  same.    What  will  be  its  pressure?     Ans.  16.73  Ibs. 

3.  A  pound  of  steam   at  temperature  of   232.8°    and  absolute 
pressure  of  22  Ibs.  per  sq.  inch  has  a  volume  of  18  cubic  feet.   What 
will  be  its  volume  at  341°  and  pressure  of  120  Ibs.? 

Ans.  3.812  cu.  ft. 


CHAPTEK  II. 
THE  APPLICATION  OF  HEAT  TO  WATER. 

27.  The  effect  of  the  application  of  heat  to  water  is  of  the  great- 
est importance  in  the  study  of  the  steam  engine. 

Eadiant  heat  is  given  off  from  hot  bodies  in  straight  lines,  and 
the  rays  of  heat  are  subject  to  the  same  laws  as  the  rays  of  light. 
As  far  as  the  generation  of  steam  in  a  boiler  is  concerned,,  the  useful 
radiation  is  confined  to  the  furnace,  the  crowns  and  sides  of  which, 
intercepting  the  rays  of  heat  from  the  burning  fuel,  become  them- 
selves heated,  and  the  heat  passes  through  them  to  the  water  in  the 
boiler.  A  considerable  amount  of  heat  is  given  off  by  radiation 
from  burning  coal,  and  it  is  very  important  to  intercept  this,  and 
to  insure,  as  far  as  possible,  that  the  whole  of  the  heat  diffused  in 
this  way  should  be  transmitted,  directly  or  indirectly,  to  the  water 
in  the  boiler,  and  not  wasted  on  the  external  air. 

Eadiation  is  an  important  item  to  be  considered  with  reference 
to  the  economical  employment  of  steam,  for  it  always  causes  a  cer- 
tain loss  of  heat,  and  unless  proper  precautions  are  taken  this  loss 
may  become  very  considerable.  The  surfaces  of  the  boilers,  steam 
pipes,  cylinders,  etc.,  when  the  engines  are  at  work,  are  very  much 
hotter  than  the  surrounding  bodies,  and  in  order  that  the  loss  of 
heat  by  radiation  may  be  avoided  as  far  as  possible,  all  these  sur- 
faces should  be  clothed,  or  lagged,  with  some  non-conducting  ma- 
terial. 

The  transfer  of  heat  by  convection  is  the  manner  in  which  liquids 
and  gases  are  heated.  When  heat  is  applied  to  the  bottom  of  a 
vessel  containing  a  fluid,  liquid  or  gaseous,  the  particles  in  contact 
with  the  bottom  are  first  heated,  and,  becoming  less  dense,  they 
therefore  rise,  allowing  cooler  particles  to  take  their  places,  which 
become  themselves  heated,  rise  and  circulate  through  the  mass  in  a 
similar  manner. 

It  is  essential  that  circulation  and  mixture  of  all  the  particles  of 
a  fluid  should  take  place  to  cause  the  temperature  to  be  uniform 
throughout  the  mass.  In  order  that  heat  may  be  efficiently  trans- 
mitted through  boiler  plates  and  flues,  each  of  the  fluids  in  contact 


12  THE  ELEMENTS  OF  STEAM  ENGINEERING 

with  them,  viz.,  the  water  on  one  side,  and  the  heated  gases  on  the 
other,  should  have  free  circulation,  so  that  the  particles  in  contact 
with  the  plates  should  not  be  considerably  different  in  temperature 
from  those  at  some  distance  from  the  plate.  Boilers  are  some- 
times fitted  with  circulating  plates  to  set  up  currents  in  the  water, 
and  with  bafflers  in  the  flues  to  break  up  the  currents  of  hot  gases 
and  form  eddies,  in  order  to  promote  circulation  and  mixture  in  the 
respective  fluids. 

In  order  to  render  most  efficient  the  transfer  of  heat  from  one 
fluid  to  another  through  a  plate,  the  general  movement  of  the  two 
fluids  should  be  in  opposite  directions,  so  that  the  hottest  parts  of 
the  two  fluids  are  opposed  to  each  other  and  the  difference  of  tem- 
peratures between  them  a  minimum.  Consequently,  in  a  boiler,, 
in  order  that  the  best  results  may  be  obtained,  the  feed  pipes  and 
circulating  plates  should  be  so  arranged  that  the  general  movement 
given  to  the  water  should  be  as  far  as  possible  in  the  opposite  direc- 
tion to  that  of  the  hot  gases  on  their  way  to  the  funnel.  For  the 
same  reason,  the  action  of  surface  condensers  is  most  efficient  when 
the  cold  water  for  condensing  the  steam  enters  the  condenser  at 
the  end  at  which  the  condensed  water  leaves  it,  so  that  the  entering 
steam  will  be  opposed  to  the  water  which  has  been  heated  to  some 
extent  by  its  passage  through  the  condenser. 

Upon  the  application  of  heat  to  the  water  in  the  boiler  the  tem- 
perature is  at  first  raised.  The  particles  of  water  in  contact  with 
the  heating  surface — which  in  a  marine  boiler  consists  of  the  fur- 
nace plates,  combustion  chamber,,  and  tubes — become  heated,  and 
rise  and  circulate  through  the  mass  of  the  water,  their  places  being 
taken  by  cooler  particles,  till  at  length  the  whole  of  the  water  is 
raised  in  temperature  to  the  boiling  point  by  the  process  of  con- 
vection. 

28.  Sensible  Heat.— The  heat  added  to  the  water  up  to  the  tem- 
perature at  which  boiling  occurs  is  called  sensible  heat,  its  effect 
being  simply  to  change  the  temperature,  and  not  the  state,  of  the 
water,  and  its  amount  or  quantity  may  easily  be  calculated;  for  it 
must  be  understood  that  the  quantity  of  heat  of  a  body,  or  the 
amount  of  heat  energy  which  a  body  gains  or  loses  in  passing 
through  a  given  range  of  temperature,  is  measured  in  thermal  units, 
commonly  denoted  by  B.  T.  U.  (British  Thermal  Units),  and  is 
given  by  the  continued  product  of  its  weight,  range  of  temperature, 
and  specific  heat. 


THE  APPLICATION  OF  HEAT  TO  WATER  13 

29.  Latent  Heat. — After  the  temperature  of  boiling  is  reached, 
and  in  order  to  convert  the  boiling  water  into  steam,  a  large  quan- 
tity of  heat  has  to  be  expended  which  does  not  produce  any  increase 
in  the  temperature.     This  heat  is  known  as  latent  heat.     It  was 
supposed  at  an  early  period  that  heat  was  a  substance,  and  that  the 
heat  thus  expended  became  hidden  or  latent  during  the  change 
from  the  liquid  to  the  gaseous  state,  and  that  it  again  became  sen- 
sible on  the  reverse  process  being  performed.     The  science  of  ther- 
modynamics, however,  has  shown  that  heat  is  not  a  substance,  but 
a  form  of  mechanical  energy,  and  that  this  heat  is  expended  in  per- 
forming the  internal  work  of  overcoming  the  molecular  cohesion 
of  the  particles  of  water  which  resist  the  change  of  state,  and  also 
performing  the  external  work  of  overcoming  the  resistance  of  ex- 
ternal bodies  to  the  change  of  volume  which  ensues.     Work  is 
therefore  done  on  changing  the  water  from  a  liquid  into  a  gas,  and 
this  is  stored  up  as  mechanical  energy,  which  can  be  yielded  back 
again  either  in  work,  or  in  heat,  when  the  gas  returns  to  its  original 
state  of  water. 

It  is  now  seen  that  this  latent  heat  of  evaporation  performs  the 
work  which  is  necessary  in  the  interior  of  the  water  and  exterior 
to  it,  the  external  work  being  a  measure  of  the  useful  effect  directly 
obtained. 

In  the  case  of  the  boiler  and  the  steam  engine,  the  external  work 
of  evaporation  is  represented  by  the  work  performed  by  the  piston 
while  the  cylinder  is  in  communication  with  the  boiler.  It  will  be 
seen  later  on  that  this  external  work  is  the  equivalent  of  but  a 
small  part  of  the  latent  heat,  the  greater  part  being  expended  in 
internal  work,  and  is  represented  by  the  energy  acquired  by  the 
water  during  the  evaporation. 

The  term  latent  heat  has  been  retained  for  convenience,  but  it 
must  be  understood  as  an  expression  that  means  simply  the  quan- 
tity of  heat  that  must  be  added  to  or  subtracted  from  a  body  in  a 
given  state,  to  change  it  into  another  state  without  altering  its  tem- 
perature. 

30.  Boiling  Point.— The  boiling  point,  or  the  temperature  of 
ebullition,  of  any  liquid  may  be  defined  .as  that  stage  in  the  addition 
of  heat  to  the  liquid  at  which  the  pressure  on  it  is  just  overcome 
by  the  pressure  of  vapor  due  to  the  temperature. 

The  temperature  of  the  boiling  point  depends  on  the  pressure 
under  which  the  liquid  is  evaporated.  The  greater  the  pressure, 


14  THE  ELEMENTS  OP  STEAM  ENGINEERING 

the  higher  is  the  temperature  at  which  the  liquid  boils.  The  tem- 
perature depends  upon  the  total,  or  absolute,  pressure,  i.  e.,  the  pres- 
sure including  that  of  the  atmosphere,  so  that  in  ascertaining  the 
temperature  corresponding  to  any  given  pressure,  as  shown  on  the 
ordinary  pressure  gauge,  the  atmospheric  pressure  must  be  added. 
The  boiling  point  of  a  liquid  is  affected  by  its  density.  Ordin- 
ary sea-water  contains  about  ^  of  its  weight  in  salt,  and  this  raises 
the  temperature  of  the  boiling  point  by  1.2°  F.,  so  that  the  boiling 
point  of  sea-water  under  the  atmospheric  pressure  is  213.2°  F. 

31.  The  Generation  of  Steam.  —  The  generation  of  steam  under 
constant  pressure  and  at  constant  volume  will  best  be  understood 
by  the  consideration  of  the  two  following  examples,  respectively. 
These  examples  are,  substantially,  as  given  in  Northcott's  "  The 
Steam  Engine,"  and  in  Sennett's  "  The  Marine  Steam  Engine." 

Example  I.  —  The  Generation  of  Steam  under  Constant  Pres- 
sure. —  Imagine  one  pound  of  water  at  a  temperature  of  32°  F.,  or 
0°  C.,  contained  in  an  upright  tube  of  indefinite  height,  and 
1  sq.  ft.,  or  144  sq.  inches,  cross  sectional  area.  The  one  pound 

i  r/oft 
of  water  has  a  volume  of  -Q^-J  =  27.7  cubic  inches  nearly,  and  it 

27  7 
will  consequently  occupy     ^r  —  0.1923  inch  of  the  height  of  the 


tube,  or  the  water  will  expose  an  area  of  surface  equal  to  one  square 
foot  and  will  be  0.1923  inch  deep.  The  tube  is  to  be  opened  to  the 
atmosphere,  so  that  each  square  inch  of  the  water  will  be  pressed 
upon  by  the  average  atmospheric  pressure  of  14.7  Ibs.,  equal  to 
14.7  X  144  =  2116.8  Ibs.  per  sq.  ft.  On  heat  being  applied  to 
this  water,  the  temperature  rises  until  212°  F.  is  reached,  when 
vaporization  commences,  and  there  is  no  further  rise  in  temperature 
until  the  whole  of  the  water  is  converted  into  steam,  the  tempera- 
ture of  which  will  also  stand  at  212°  F. 

Immediately  previous  to  the  commencement  of  vaporization  a 
quantity  of  heat  will  have  been  imparted  to  the  water  equal  to 
1(212  —  32)1  =  180  thermal  units,  or  to  180X778  =  140,040 
foot-pounds.  When  the  pound  of  water  is  all  converted  into  steam 
a  further  quantity  of  heat  will  have  been  imparted  equal  to  965.7 
thermal  units,  or  to  965.7  X  778  =  751,315  foot-pounds;  and  yet, 
notwithstanding  this  large  absorption  of  heat,  the  temperature 
remains  at  212°  F.  Since  this  heat  has  changed  the  sitate  of  the 
water  from  that  of  a  liquid  to  that  of  a  gas  without  raising  its 


THE  APPLICATION  OF  HEAT  TO  WATER  15 

temperature,  it  is,  from  our  definition,  latent.  We  know,  however, 
that  the-  only  manner  in  which  heat  energy  can  disappear  as  heat, 
is  by  its  being  transformed  into  some  other  form  of  energy  or  in 
the  performance  of  work;  and,  in  order  to  understand  what  has 
become  of  this  large  quantity  of  heat,  we  must  inquire  as  to  what 
work  has  been  done. 

In  the  first  place  180  units  were  absorbed  in  raising  the  tempera- 
ture of  the  water  from  32°  to  212°,  and  this  we  know  as  the  sensible 
heat.  This  change  of  temperature  is  practically  the  only  change 
effected  up  to  that  point  of  the  process,  as  the  expansion  of  the 
water  in  rising  from  32°  F.  to  212°  F.,  is  extremely  small  and  may 
be  disregarded.  The  effect  of  the  heat  abstracted  from  the  source  of 
heat  and  communicated  to  the  water  up  to  this  point  has  been  only 
to  raise  the  temperature  of  the  water. 

Steam  of  a  pressure  of  14.7  Ibs.  per  sq.  inch  has  been  found  to 
occupy  a  volume  1644  times  the  water  from  which  it  was  gener- 
ated; that  is,  a  cubic  inch  of  water  is  converted  into  1644  cubic 
inches  of  steam  at  atmospheric  pressure  of  14.7  Ibs.  per  sq.  inch. 
The  steam  from  one  pound  of  water  will  thus  have  a  volume  of 

1644  X  27  7 

..yog =  26.36   cubic   feet.     In   expanding   to   this   volume 

under  atmospheric  pressure,  it  will  necessarily  have  performed  ex- 
ternal work  to  the  extent  of  26.36  X  2116.8  =  55,798  foot-pounds. 
This,  perhaps,  may  be  more  clearly  seen  if  we  disregard  the  pres- 
sure of  the  atmosphere,  and  assume  that  the  steam  expands  in  the 
tube  under  a  piston  weighing  14.7  X  144  =  2116.8  Ibs.,  which  it 
will  evidently  have  to  lift  in  the  tube  through  a  height  of  26.36 
feet. 

The  heat  required  to  convert  the  water  at  212°  F.  into  steam  of 
the  same  temperature  has  been  found  by  experiment  to  be  965.7 
units,  or  965.7  X  778  =  751,315  foot-pounds,  and  of  this,  55,798 
foot-pounds  are  accounted  for  in  the  external  work  as  shown  above, 
leaving  751,315  —  55,798  =  695,517  foot-pounds  as  the  mechani- 
cal equivalent  of  the  internal  work,  or  of  the  heat  absorbed  in 
effecting  the  expansion  against  the  internal  or  molecular  forces. 

Of  the  total   quantity   of  heat  energy  imparted   to   the  water 

— ipjo — —  180  thermal  units  have  been  expended  in  increasing  the 

temperature  of  the  water  by  180°  ;  — ~~  —  =  893.98  thermal  units 
have  been  expended  in  the  internal  work  of  expanding  the  steam 


16  THE  ELEMENTS  OF  STEAM  ENGINEERING 

55  798 
against  molecular  forces,  and      ^    =71.72  thermal  units  have 

been  expended  in  the  external  work  of  overcoming  the  resistance  of 
the  atmosphere  which  opposed  the  expansion  of  volume.  The  sum 
of  these  three  quantities  of  heat,  that  is,  the  sum  of  the  units  due 
to  the  sensible  heat,  the  internal  work,  and  the  external  work,  is 
known  as  the  total  heat  of  steam,  or  the  total  heat  of  vaporization, 
and  is,  in  this  case,  180  +  893.98  +  71.72  =  1145.7  thermal  units, 
which  is  the  total  heat  of  one  pound  of  steam  at  temperature  of 
212°  F. ;  that  is,  it  is  the  units  of  heat  required  to  evaporate  one 
pound  of  water  from  32°  F.  into  steam  at  212°  F. 

The  results  of  very  careful  experiments  have  shown  that  the  total 
heat  of  steam  increases  but  slightly  as  the  temperature  increases, 
the  increment  of  increase  being  0.305  of  a  thermal  unit  for  each 
degree  of  increase  in  temperature.  This  fact  enables  us  to  deduce 
a  formula  for  ascertaining  the  total  heat  of  steam  of  any  tempera- 
ture. Thus,  knowing  the  total  heat  of  steam  at  212°  F.  to  be 
(accurately)  1146.6  units,  we  will  have  for  the  total  heat,  HT,  at 
any  other  temperature  t,  reckoning  from  32°,  the  formula: 

HT—  1146.6  +  .305(^  —  212). 

It  will  be  convenient  to  change  the  form  of  this  expression,  as 
follows : 

#2,  =,1146.6  -f  .305(£  —  212)  =  1146.6  +  .305  [t  —  (32  +  180)] 
=  1091.7 +  .305(J  — 32). 

The  sum  of  the  two  quantities  of  heat  that  disappears,  represent- 
ing the  internal  and  external  work,  is  equal  to  the  total  heat  minus 
the  sensible  heat,  above  32°  F.,  and  is  known  as  the  latent  heat  of 
vaporization,  or  the  latent  heat  of  steam  under  the  given  pressure, 
and  equals  in  this  case  893.98  +  71.72  =  965.7  thermal  units. 

It  has  been  experimentally  determined  that  the  increment  of 
decrease  in  the  latent  heat  of  steam  is  0.7  of  a  thermal  unit  for 
each  degree  of  increase  of  temperature.  We  can,  therefore,  deduce 
a  formula  for  ascertaining  the  latent  heat  of  steam  at  any  tem- 
perature. 

Thus,  knowing  the  latent  heat  of  steam  at  212°  F.  to  be  965.7 
units  we  will  have  for  the  latent  heat,  H L,  at  any  temperature  t, 
reckoning  from  32°  F.,  HL  =  965.7  —.7  (t  —  212°).  We  may 
change  the  form  of  this  formula  in  order  to  make  it  very  similar  to 
that  for  the  total  heat,  and  consequently  more  easily  remembered. 


THE  APPLICATION  OF  HEAT  TO  WATER 


17 


Thus,  HL  =  965.7  —  .7(*  —  212)  =  965.7  —  .7  [t  —  (32  +  180) ] 

=  1091.7  —  .7(^  —  32). 

An  important  modification  in  the  conditions  under  which  the 
steam  is  generated  in  the  above  example  will  now  be  considered. 

Example  II. — The  Generation  of  Steam  at  Constant  Volume. — 
In  Example  I,  the  piston  was  free  to  move  and  allowed  the  steam 
to  gradually  expand  in  volume  as  the  water  evaporated.  We  will 
now  suppose  this  not  to  be  the  case,  and  the  piston  is  fixed  in  the 
cylinder  at  a  certain  distance  above  the  water  as  shown  in  Fig.  1. 
Suppose  further  that  only  water  is  contained  in  the  space  A B,  so 
that  there  is  no  pressure  therein.  If  heat  be  now  applied  to  the 
water,  instead  of  the  formation  of  steam  taking  place  only  when  a 
certain  temperature  is  reached,  as  in  our  first  case,  steam  imme- 


1. 

diately  commences  to  form,  and  both  the  temperature  and  pres- 
sure gradually  increase  as  more  of  the  water  is  evaporated,  although 
the  temperature  and  pressure  are  still  connected  by  the  same  law  as 
before. 

The  pressure  and  temperature  of  the  steam  when  all  the  water  has 
just  been  evaporated  can  be  ascertained  from  our  knowledge  of  the 
weight  of  water,  and  the  known  volume  of  the  steam;  for  the 
weight  of  the  steam  is  the  same  as  that  of  the  water  from  which  it 
was  formed,  so  that  its  volume  per  pound  can  be  calculated  and  its 
pressure  and  temperature  then  ascertained  from  the  tables. 

It  will  be  seen  that  this  can  occur  in  practice  when  steam  is  being 
raised  in  a  boiler  from  cold  water  with  the  stop  valve  and  other  out- 
lets closed,  except  that  instead  of  there  not  being  any  pressure  on 
the  water  prior  to  the  application  of  heat,  there  is  the  pressure  of 


18  THE  ELEMENTS  OF  STEAM  'ENGINEERING 

the  atmosphere.  In  this  case  the  temperature  of  the  water  will 
gradually  rise,  but  no  steam  will  be  formed,  and  the  pressure  will 
not  increase,  until  the  temperature  reaches  that  corresponding  to 
the  atmospheric  pressure,  viz.,  212°  F.,  when  steam  commences  to 
be  given  off,  and  both  temperature  and  pressure  rise.  It  is  thus 
seen  that  the  absolute  pressure  of  boiler  steam  is  in  excess  of  the 
gauge  pressure  by  an  amount  equal  to  the  atmospheric  pressure.  In 
our  practical  illustration,  also,  the  evaporation  of  the  water  does 
not  continue  until  all  the  water  is  evaporated,  as  the  amount  of 
water  is  so  large  compared  with  the  volume  of  the  boiler — from 
65  per  cent  to  75  per  cent  of  the  volume  of  the  boiler — that  the 
desired  pressure  is  obtained  when  only  a  small  portion  of  the  water 
has  been  evaporated. 

31a.  Formula  Connecting  Pressure  and  Temperature. — The  rela- 
tions between  the  pressure  and  temperature  of  steam  are  very  com- 
plicated, and  for  practical  purposes  these  relations  are  best  ascer- 
tained from  tables  giving  the  properties  of  saturated  steam.  Many 
formulas  have  been  devised  to  represent  the  relations,  but  most  of 
them  are  of  only  theoretical  interest,  and  those  of  logarithmic  form 
the  most  accurate. 

Eankine  gives  the  following  equation  connecting  the  temperature 
and  pressure  of  saturated  steam,  which  gives  fairly  accurate  results : 

B  C 

log  p  =  < 4—  7,— 7p2, 

in  which  T  =  absolute  temperature  =  t  -\-  461.2°.  For  a  value  of 
p  in  pounds  per  square  inch  the  values  of  A,  B,  and  C  are :  A  =• 
6.1007,  log  B  =  3.43642,  and  log  C  =  5.59873. 

EXAMPLE. — Find  the  pressure  per  gauge  of  steam  having  a  tem- 
perature of  285°  F. 

Solution:  log  p  —  A  —  ^  —  ~2.        T  —  285  -f  461.2  =  746.2. 

746.2  log  2.87286 
746.2  log  2.87286 

log  5.74572 
C  log  5.59873 

-^=0.71287  log  9.85301  —  10 


THE  APPLICATION  OF  HEAT  TO  WATER  19 

B  log  3.43642 
746.2  log  2.87286 

B  =  3.66067  log  0.56356 
1  A  =  6.10070 

:P  +  ^_2—  3.66067  +  0.71287  =  4.37354 

p  —  53.35  log  1.72716 
Pressure  per  gauge  =  53.35  — 14.7  =  38.65  Ibs.  per  sq.  inch. 

EXAMPLE. — Find  the  temperature  on  the  Fahrenheit  scale  of 
steam  of  80  Ibs.  absolute  pressure  per  square  inch. 
Solution:     Solving  for  T,  we  have : 
T_B 

A  —  6.10070 
p  =  80  log  1.90309 

A  —  log  p  —  4.19761  log  0.62300 
C  log  5.59873  m 
4  log  0.60206  ' 

40 (A— log  p)=  6664900  log  6.82379 

B  log  3.43642 
B  log  3.43642 

B2—    7461700  log  6.87284 

14126600  log  7.15003,  £  log  3.57502  antilog  =  3758.6 


Therefore,  ^±C(A  —  logp)+B2  =  3758.6 

B  =2731.6  log  3.43642 

6490.2 log  3.81225 

A  —  logp  —  4.19761  log  0.62300  colog,  9.37700  —  10 
2  log  0.30103  colog  9.69897  —  10 

T—  773.1  log  2.88822 
2  =  773.1  — 461.2  =  311.9°. 

32.  Formula  Connecting  Pressure  and  Volume. — Various  formu- 
las have  been  devised  to  show  the  relation  between  the  pressure  and 
volume  of  saturated  steam. 

The  following  formula  gives  results  that  are  quite  accurate  within 
the  range  of  pressures  of  recent  practice,  viz.,  100  Ibs.  to  280  Ibs. 
per  sq.  inch : 

.    pvU  =482, 


20  THE  ELEMENTS  OF  STEAM  ENGINEERING 

in  which  p  =  absolute  pressure  in  pounds  per  sq.  inch,  and  v  =  vol- 
ume of  one  pound  of  steam  in  cubic  feet  at  pressure  p. 

EXAMPLE. — Find  volume  of  a  pound  of  steam  at  280  pounds  ab- 
solute pressure. 

AR9 

Solution :     pv&  =  482,  whence  i;«  =  S£ 

/ooO 

U  log  v  =  log  482  -f  colog  280 
482  log- 2.68305 
280  colog  7.55284  — 10 

log  0.23589  Hog  9.37271  —  10 

16  log  1.20412 
17  colog  8.76955  —  10 

v  —  1.6675  log  0.22207  Hog  9.34638  —  10. 

EXAMPLE. — Find  volume  of  a  pound  of  steam  at  500  Ibs.  abso- 
lute pressure. 

Solution:  pv&  =  482,  whence  -J-J  log  v  =  log  482  —  log  500 
482  log  2.68305 
500  log  2.69897 

log  9.98408  — 10 
log  (— )  0.01592  Hog  (— )  8.20194  —  10 

16  log  1.20412 

17  colog  8.76955  —  10. 

-  0.01498  Hog  (— )  8.17561  —  10 
v  =,  0.9661  log  9.98502  —  10. 

PEOBLEMS. 

4.  The  specific  heat  of  mercury  is  0.033.     How  many  pounds  of 
it  at  a  temperature  of  240°  will  be  necessary  to  raise  the  tempera- 
ture of  12  pounds  of  water  from  50°  to  58°  ?  Ans.  16  Ibs. 

5.  Work  at  the  rate  of  a  horse-power  is  expended  for  an  hour  in 
creating  friction,  the  heat  from  which  is  all  communicated  to  10 
cubic  feet  of  water  contained  in  a  non-conducting  tank..    The  orig- 
inal temperature  of  the  water  was  60°.     What  will  be  its  tempera- 
ture at  the  end  of  the  hour?  Ans.  64.1°. 

6.  The  formula  connecting  the  temperature  and  pressure  of  steam 

B       C 
is  log  p  =  A  —  -ff,  —  -fpi ,  in  which  p  =  absolute  pressure  in  pounds 


THE  APPLICATION  OF  HEAT  TO  WATER  21 

per  square  inch,  4  =  6.1007,  log  B  =  3.43642,  log  C  =  5.59873, 
and  T  =  the  absolute  temperature.  Find  the  temperature  of  steam 
of  120  Ibs.  absolute  pressure.  Ans.  341.69°. 

7.  The  formula  connecting  the  pressure  and  volume  of  steam  is 
pv™  =  482,  in  which  p  =  absolute  pressure  in  pounds  per  square 
inch,  and  v  =  volume  in  cubic  feet  of  a  pound  of  steam  at  pressure 
p.  Find  the  volume  of  one  pound  of  steam  at  pressure  of  515  Ibs. 

Ans.  0.9396  cubic  feet. 


CHAPTEE  III. 
FUELS  AND  COMBUSTION. 

33.  The  fuels  commonly  used  for  generating  steam  are  coal, 
coke,  wood,  and  mineral  oil. 

Coal  is  an  insoluble  combustible  mineral  widely  distributed  over 
the  earth,  and  of  incalculable  value  in  the  application  of  the 
sciences.  More  than  1,000,000,000  tons  <are  mined  yearly,  and  it 
is  estimated  that  there  is  enough  in  the  crust  of  the  earth  to  make 
a  layer  3  feet  thick  over  the  land  surface. 

Coal  consists  essentially  of  carbon,  together  with  hydrogen,  oxy- 
gen, nitrogen,  and  sulphur,  and  its  formation  is  the  result  of  a  dis- 
tillation of  organic  matter  in  the  earth,  the  process  being  conducted 
on  an  immense  scale  under  the  influence  of  heat  and  pressure,  and 
for  an  unlimited  time.  If  the  distillation  is  complete  the  result  is 
pure  carbon  in  the  shape  of  graphite.  If  less  complete  the  result  is 
hard  coal,  called  anthracite,  which  may  contain  more  than  90  per 
cent  of  carbon.  If  still  less  complete,  the  result  is  soft,  or  bitu- 
minous, coal,  containing  upwards  of  60  per  cent  of  carbon. 

34.  Anthracite  Coal  is  hard  and  lustrous,  makes  an  intense  fire,, 
and  when  dry  burns  without  smoke  and  with  very  little  flame.     It 
contains  but  a  small  percentage  of  volatile  matter.     It  is  much 
more  expensive  and  far  less  abundant  than  soft  coal. 

35.  Bituminous  or  Soft  Coal  contains  less  carbon  than  anthracite, 
but  is  richer  in  hydrogen  and  has  a  larger  percentage  of  volatile 
matter  which  is  driven  off  when  the  coal  is  ignited.     Bituminous 
coal  contains  no  bitumen,  its  name  being  due  to  the  fact  that  its 
behavior  when  heated  is  similar  to  that  of  bitumen — that  is,  it 
swells  more  or  less,  fuses,  and  burns  with  a  bright  flame  and  con- 
siderable dense  smoke. 

Varieties  of  coal  known  as  semi-anthracite  and  semi-bituminous 
are  so  called  from  their  near  approach  in  character  to  anthracite  or 
bituminous  coal. 

36.  Coke  is  the  solid  product  resulting  from  the  destructive  dis- 
tillation of  bituminous  coal,  and  has  more  value  for  its  use  in  the 
metallurgic  process  of  reducing  iron  than  for  the  purpose  of  gen- 


FUELS  AND  COMBUSTION  23 

erating  steam.  Coke  is  carbonized  bituminous  coal,  the  volatile 
constituents  having  been  driven  off  by  the  application  of  heat,  and 
it  bears  to  coal  the  same  relation  that  charcoal  bears  to  wood. 

37.  Wood  is  only  used  for  generating  steam  in  regions  where  it  is 
abundant,  or  where  coal  is  scarce  or  difficult  to  obtain.     The  rela- 
tive value  of  the  heat  producing  qualities  of  wood  and  coal  is  ex- 
pressed by  assigning  to  a  cord  of  hard  wood  the  equivalence  of  a 
ton  of  anthracite  coal. 

38.  Mineral  Oil  as  a  fuel  is  destined  in  the  future  for  extensive 
use,  but  at  present  the  imperfection  of  the  method  of  firing  it,  its 
unequal  distribution  over  the  earth,  and  the  danger  attending  its 
storage  and  carriage,  has  limited  its  application  for  steam  generating 
purposes.     The  oil  used  for  this  purpose  is  crude  petroleum  divested 
by  distillation  of  its  most  volatile  and  dangerous  parts.     The  usual 
arrangement  for  firing  is  to  spray  the  oil  into  the  furnace  by  means 
of  a  steam  jet,  which  also  induces  a  sufficient  supply  of  air  for  com- 
bustion.   An  ordinary  fire  is  first  kindled  on  the  grate  to  produce 
steam  of  sufficient  pressure  for  the  jet,  after  which  oil  only  is  sup- 
plied as  fuel.     The  richness  in  hydrogen  of  this  crude  petroleum 
gives  it  an  evaporative  effect  of  1.5  times  that  of  coal. 

39.  Owing  to  its  extreme  abundance  coal  is  the  fuel  generally  used 
in  the  furnaces  of  boilers,  and  the  processes  through  which  it  passes 
during  combustion  should  be  studied  in  order  that  precautions  may 
be  taken  to  insure  economy  in  its  use. 

40.  Carbon  and  hydrogen  are  the  heat-producing  elements  of  coal, 
the  small  quantity  of  sulphur  ever  present  contributing  very  little 
heat.     The  presence  of  oxygen  in  a  fuel  really  detracts  from  the 
total  heat  of  combustion,  for  it  is  known  to  combine  with  one-eighth 
of  its  weight  of  hydrogen  to  form  water — existing  as  such  in, the 
fuel — and  only  such  hydrogen  in  excess  of  one-eighth  of  the  oxygen 
may  be  reckoned  in  estimating  the  total  heat  of  combustion  of  a 
fuel. 

41.  Combustion  in  a  furnace  is  a  rapid  oxidation  caused  by  the 
chemical  union  of  the  oxygen  of  the  air  with  the  hydrogen  and 
carbon  of  the  coal,  heat  being  produced  in  the  process.    It  has  been 
seen  that  coal  consists  largely  of  carbon,  part  of  which  is  free,  while 
the  other  part  is  in  combination  with  hydrogen,  forming  hydro- 
carbons.    These  hydro-carbons  are  solid  at  ordinary  temperatures, 


24  THE  ELEMENTS  OF  STEAM  ENGINEERING 

but  being  volatile  they  take  the  form  of  gas  when  heated  to  tem- 
peratures below  that  necessary  for  combustion.  The  gases  thus 
formed  consist  of  compounds  of  hydrogen  and  carbon  which  readily 
decompose  under  the  action  of  heat,  the  carbon  becoming  more  dis- 
tinctly separated  from  the  compounds  as  the  temperature  rises. 
When  under  the  influence  of  temperature  sufficiently  high,  which 
not  infrequently  occurs  in  furnaces,  the  hydrogen  and  carbon  be- 
come entirely  separated,  and  as  each  of  these  elements  has  a  stronger 
affinity  for  oxygen  than  for  each  other  they  will,  if  the  supply  of  air 
be  just  sufficient,  combine  with  oxygen  separately  and  burning  en- 
sues. Under  such  ideal  conditions  combustion  would  be  perfect, 
with  no  formations  of  soot  or  smoke,  the  results  being  carbonic  acid 
(C02),  superheated  steam  in  process  of  falling  to  water,  and  free 
nitrogen.  The  earthy  impurities,  amounting  to  from  2  per  cent  to 
10  per  cent,  would  form  the  residue  of  ash  and  clinker. 

If  the  supply  of  air  through  the  ash-pit  be  insufficient  for  com- 
plete combustion,  or  if  it  is  not  perfectly  mixed  with  the  gases  the 
result  of  the  union  of  the  carbon  and  oxygen  will  be  carbonic  oxide 
(CO),  which  produces  less  than  one-third  the  heat  due  the  produc- 
tion of  C02 ;  and  unless  provision  is  made  for  the  admission  of  air 
above  the  fuel  to  supply  oxygen  to  convert  this  CO  into  C02,  much 
of  the  carbon  escapes  unconsumed,  resulting  in  a  loss  of  fuel,  the 
production  of  smoke,  and  the  deposit  of  soot  in  the  tubes  and  up- 
takes. The  prevention  of  smoke  is  only  possible  when  the  supply 
of  air  to  the  gases  is  sufficient  when  they  are  at  the  temperature  of 
ignition,  and  as  the  gases  after  formation  pass  off  very  quickly 
from  the  furnace,  it  has  been  found  to  be  practically  necessary  to 
supply  one  and  a  half  times  the  quantity  of  air  theoretically  neces- 
sary for  combustion  if  the  draft  be  artificial,  and  twice  the  theoreti- 
cal amount  if  the  draft  be  natural. 

42.  The  quantity  of  air  necessary  for  the  combustion  of  a  pound 
of  carbon  may  be  estimated  as  follows:  In  the  formation  of  CO, 
the  C  and  0  combine  by  weight  in  the  proportion  of  12  to  32,  or  as 
1  to  2f ;  it  follows  that  every  pound  of  C  requires  2f  Ibs.  of  0. 
This  0  must  come  from  the  air  containing  oxygen  and  nitrogen  in 
the  proportion  of  23  parts  of  0  and  77  parts  of  N"  in  every  100  parts 
by  weight.  Since  1  cubic  foot  of  air  weighs  very  nearly  0.08  lb., 
it  contains  0.23  X  0.08  =  0.0184  lb.  of  0.  To  produce  2f  Ibs.  0, 

the  amount  required  for  the  combustion  of  one  pound  of  carbon,  we 

02 
must,  therefore,  have   -foT  —  145  cubic  feet  of  air. 


FUELS  AND  COMBUSTION  25 

Eankine  gives  the  following  formula  for  the  weight  of  air  theo- 
retically necessary  for  the  complete  combustion  of  a  pound  of  fuel : 


where  A  =.  pounds  of  air  required  per  pound  of  the  fuel,  and  (7,  H, 
and  0  are  the  fractions  of  carbon,  hydrogen,  and  oxygen,  respect- 
ively, contained  in  a  pound  of  the  fuel. 

This  formula  is  obtained  from  the  following  considerations: 
The  average  composition  of  air  is,  by  weight,  as  we  have  seen,  77 
parts  of  nitrogen  and  23  parts  of  oxygen.     For  the  complete  com- 
bustion of  one  pound  of   carbon    (C02)    there  will   be  required 

DO 

-£  =  2f  pounds  oxygen,  and  since  each  pound  of  air  contains,  but 

0.23  pound  oxygen,  there  will  be  required  for  the  combustion  of 

22 

one  pound  of  carbon  ~  Jo  =11.6  pounds  air.     For  the  complete 

combustion  of  one  pound  of  H  there  are  required  8  pounds  0. 

This  0  must  be  abstracted  from  the  air,  and  there  will  be  required, 

g 
therefore,  ^-^  =35  pounds  air  to  furnish  it,  which  is  three  times 

as  much  as  was  required  for  the  combustion  of  the  pound  of  carbon. 
In  estimating  the  thermal  value  of  a  fuel  only  such  H  as  is  unbal- 
anced by  0  is  taken  as  available  for  the  generation  of  heat,  and 

therefore  H  —  -^-  is  the  expression  for  this  free  H.     The  H  in  the 

coal  is  diminished  by  one-eighth  of  the  0  because  in  the  formation 
of  the  coal  by  natural  processes  that  amount  has  already  combined 
with  0  to  form  water — existing  as  such  in  the  fuel — and  is  there- 
fore no  longer  available  for  generating  heat.  In  the  formula  given, 
11.6  is  replaced  by  the  approximation  12,  and  all  within  the  large 
brackets  is  expressed  in  terms  of  carbon. 

The  amounts  of  heat  liberated  by  the  combustion  of  a  given 
quantity  of  carbon  and  of  hydrogen  have  been  determined  experi- 
mentally, and  the  results  are  as  follows : 

A  pound  of  H  on  entering  into  combination  with  8  pounds  0, 
liberates  a  quantity  of  heat  equal  to  about  62,000  thermal  units. 

A  pound  of  carbon  on  combining  with  2f  pounds  of  oxygen,  and 
falling  to  CO 2,  liberates  about  14,500  thermal  units. 

A  pound  of  carbon,  combining  with  1J  pounds  of  gaseous  oxygen 
to  form  CO,  generates  only  about  4500  thermal  units. 


26  THE  ELEMENTS  OF  STEAM  ENGINEERING 

43.  Total  Heat  of  Combustion  of  Fuels. — There  are  two  methods 
of  ascertaining  the  calorific  value  of  fuels :  First  by  burning  samples 
of  them  in  a  calorimeter;    and  secondly  by  chemically  analyzing 
them  and  calculating  the  heating  value  of  the  separate  components. 
The  first  plan  is  now  the  more  accurate  and  reliable. 

To  make  a  test  by  means  of  the  bomb  calorimeter  a  known  weight 
of  the  fuel  is  powdered  and  burned  in  pure  oxygen  under  pressure, 
the  containing  vessel  being  of  iron  and  submerged  in  water.  The 
ignition  is  caused  electrically,  and  the  rise  in  temperature  of  the 
water  gives  the  heating  value  of  the  fuel. 

The  method  based  on  chemical  analysis  is  not  very  reliable  be- 
cause of  the  assumption  made  that  the  value  of  the  elements  as 
existing  in  the  fuel  is  the  same  as  if  they  existed  in  the  separate 
and  uncombined  condition,  and  that  no  heat  is  -absorbed  in  the 
separation  of  the  carbon  and  hydrogen  of  the  hydro-carbons.  In 
the  absence  of  exact  formula?,  the  calorific  value  of  a  fuel  may  be 
very  approximately  determined  by  the  following  formula : 

h  =. 14,500 ["0  +  4.2S(H-  °\] , 
L  V          o  /  J 

where  h  =  the  total  heat  of  combustion  of  a  pound  of  the  fuel  in 
thermal  units,  and  (7,  H,  and  0  are  the  fractional  weights  of  car- 
bon, hydrogen,  and  oxygen  in  the  fuel.  As  already  explained,  the 

expression  f  H  -  -  -g-J  is  the  free  hydrogen,  or  the  H  not  combined 

nn  f)00 

with  0  to  form  H20.     The  thermal  value  of  this  free  H  is  J^QQ 

=  !4.28  times  the  value  of  an  equal  weight  of  carbon,  therefore  the 
whole  quantity  within  the  large  brackets  is  the  equivalent  weight 
of  the  pound  of  fuel  in  pure  carbon,  the  coefficient  14,500  being  the 
thermal  value  of  one  pound  of  carbon  as  it  exists  in  fuel. 

44.  Evaporative  Power  of  Fuel. — To  convert  one  pound  of  water 
at  a  temperature  of  212°  F.  into  steam  of  the  same  temperature 
requires,  as  was  shown,  965.7  thermal  units,  or  in  round  numbers 
966  units;  therefore,  by  dividing  the  total  heat  of  combustion  of  a 
pound  of  fuel  by  966,  the  quotient  will  be  the  number  of  pounds  of 
water  the  pound  of  fuel  is  theoretically  capable  of  converting  into 
steam  of  212°  from  water  of  212°,  and  is  equal  to 


FUELS  AND  COMBUSTION  2T 

EXAMPLE  I.— A  variety  of  good  steaming  coal  contains  90  per 
cent  by  weight  of  C,  3.5  per  cent  of  H,  6  per  cent  of  0.  Find  the 
total  heat  of  combustion,  and  the  evaporative  power  of  a  pound  of 
the  coal. 

Solution: 

Total  heat  of  combustion  =  14,500    fo.9  +  4.28  (0.055  -    ^  )  1 

—  14,757  units, 
therefore  the  evaporative  power  is       '„,.      =  15.27  pounds  water. 

We  infer  from  this  that  the  complete  combustion  of  a  pound  of 
good  coal  should  evaporate  about  15  pounds  of  water,  but  owing  to 
several  sources  of  waste  an  evaporation  of  from  10  to  11  pounds 
from  and  at  212°  F.  is  a  result  which  can  only  be  obtained  under 
the  most  favorable  circumstances.  The  principal  losses  which  con- 
tribute to  this  result  are  the  waste  of  unburned  fuel,  both  in  the 
solid  and  gaseous  state,  the  loss  of  heat  in  the  gases  escaping  through 
the  smoke  pipe,  and  the  loss  from  radiation. 

Example  in  the  Combustion  of  Coal. — Coal  containing  90  per  cent 
of  carbon,  4  per  cent  of  hydrogen,  4  per  cent  of  oxygen,  and  2  per 
cent  of  ash  is  burned  in  a  furnace  with  a  supply  of  air  of  24  pounds 
per  pound  of  coal.  Temperature  of  air  supplied  to  furnace  60 °, 
and  temperature  of  chimney  gases  600°.  The  specific  heats  of 
N,  0,  C02,  and  superheated  steam  are  respectively  0.244,  0.218, 
0.217,  and  0.475.  Find  the  temperature  of  the  furnace. 

The  combining  weights  of  0  and  N  in  air  are  as  23  to  77,  there- 
fore in  24  Ibs.  of  air  there  are  24  X  0.23  =  5.52  Ibs.  of  0,  leaving 
24  —  5.52  =  18.48  Ibs.  of  N  free.  There  are  then  5.52  +  0.04  — 
5.56  Ibs.  of  0  in  all. 

C  and  0  combine  to  form  C02  in  the  proportion  by  weight  of 
12  to  32,  or  as  1  to  2f ;  therefore,  2f  X  0.9  =  2.4  Ibs.  0  required  to 
unite  with  the  0.9  Ib.  C  in  the  coal  to  form  2.4  -f  0.9  =  3.3  Ibs. 
C02.  H  and  0  combine  to  form  water  in  the  proportion  of  2  to  16, 
or  as  1  to  8 ;  therefore,  0.04  X  8  =  0.32  Ib.  0  unites  with  the  0.04 
Ib.  of  H  in  the  coal  to  form  0.04  -j-  0.32  =  0.36  Ib.  superheated 
steam. 

There  are  2.4  +  0.32  =  2.72  Ibs.  of  0  used  altogether,  leaving 
5.56  _  2.72  =  2.84  Ibs.  of  free  0. 


28  THE  ELEMENTS  OF  STEAM  ENGINEERING 

The  gases  to  be  heated  are :  The  gases  put  in  the  furnace  were : 

18.48  Ibs.  N  24.00  Ibs.  air 

3.30     "     C02  0.90     "     C 

0.36     "     H20  0.04     "     H 

2.84     "     0  0.04     "     0 


24.98  Ibs.  gases.  24.98  Ibs.  gases.' 

Since  the  amount  of  heat  the  gases  gained  in  having  their 
temperature  raised  is  the  continued  product  of  their  temperature, 
weights,  and  specific  heats,  we  have : 

18.48  X  0.244  =  4.509  units 

3.30  X  0.217  =  0.716  " 

0.36  X  0.475  =  0.171  " 

2.84  X  0.218  =  0.619  " 

6.015  units 
required  to  raise  the  temperature  of  the  gases  1°. 

The  total  heat  of  combustion  of  a  pound  of  the  coal  is 

14,500[~0.9  +  4.28  f  0.04  -  -  ^  \1  —  15^22  thermal  units.    There- 
fore, the  elevation  of  temperature  of  the  gases  above  their  initial 

15  222 
temperature  is      '    ^-  =.  2530°,  and  the  temperature  of  the  furnace 

is  2530 +  60  =  2590°. 

EXAMPLE  II. — Suppose,  in  the  above  example,  it  were  required  to 
find  the  number  of  thermal  units  available  for  generating  steam. 
We  would  then  proceed  as  follows :  For  each  degree  of  temperature 
of  the  discharged  gases  there  are  6.015  units  of  heat  carried  off,  or 
a  total  of  (600  —  60)6.015  =  3248  units  carried  off  to  waste,  leav- 
ing 15,222  —  3248  =  11,974  units  available  per  pound  of  coal  for 
generating  steam. 

It  has  been  shown  that  the  expression  1091.7  +  0.305(2  —  32) 
gives  the  total  heat  of  a  pound  of  steam  at  temperature  t,  or,  in 
other  words,  it  is  the  total  heat  required  to  evaporate  a  pound  of 
water  from  a  temperature  of  32°  into  steam  at  temperature  t.  A' 
feed  water  temperature  of  32°  is  taken  as  an  origin,  and  the 
evaporation  is  said  to  be  from  32°  and  at  t. 

If  our  feed  water  be  taken  at  any  other  temperature  denoted  by 
tf9  we  will  have  in  every  pound  of  feed  water  l(tf  —  32)1  thermal 
units  which  will  not  be  supplied  by  the  fuel.  (The  two  factors  of 


FUELS  AND  COMBUSTION  29 

unity  represent,  respectively,  the  weight  of  water  considered  and 
its  specific  heat,  but  hereafter  they  will  be  neglected.) 

Obviously,  when  the  temperature  of  the  feed  is  tf)  an  amount 
of  heat  equal  to  tf  —  32  units  must  be  subtracted  from  the  amount 
given  by  the  formula  in  order  to  obtain  the  heat  units  to  be  sup- 
plied by  the  fuel  to  evaporate  a  pound  of  water  of  temperature  if 
into  steam  of  temperature  t. 

If,  then,  we  denote  by  Hw  the  units  of  heat  necessary  to  evapo- 
rate a  pound  of  water  from  a  temperature  tf,  and  at  a  temperature 
t,  we  will  have  the  expression,  general  ^in  its  application, 

Hw  =  1091.7  +  0.305(^  —  32)  —  (tf  —  32). 

If  t  and  tf  each  equal  212°,  we  shall  have: 

Hw  =  1091.7  +  0.305(212  —  32)  —  (212  —  32)  =  966.6  units, 
or  more  accurately  by  experiment,  965.7  units,  which,  we  have 
already  seen,  is  the  amount  of  heat  required  to  convert  a  pound  of 
water,  having  a  temperature  of  212°,  into  steam  at  the  same  tem- 
perature. 

In  the  example  above  considered  we  have  11,974  units  available 
per  pound  of  coal  for  generating  steam.  Suppose  the  feed  water  to 
be  at  a  temperature  of  32°,  and  the  water  converted  into  steam  of 
212°  temperature;  how  many  pounds  of  water  would  a  pound  of 
this  coal  evaporate  ? 

Here  we  have  Hw  =  1091.7  +  0.305(212  —  32)  —  (32  —  32)  = 

1146.6  units  required  for  each  pound  evaporated;  hence  ^  ' '  = 
10.44  pounds  of  water  evaporated  per  pound  of  coal. 

This  result  might  have  been  obtained  as  follows :  Latent  heat  of 
steam  at  212°  =  1091.7  —  0.7  (212  —  32)  =  965.7  units,  and  212  — 
32  —  180  units  required  to  raise  each  pound  of  water  from  32°  to 
212°.  965.7  +  180  =  1145.7  units  required  from  fuel  per  pound 

11  974 

of  water  evaporated ;  hence  ^^  ^  =  10.44  pounds  water  evapo- 
rated per  pound  of  coal,  as  before. 

Suppose,  in  the  above  example,  the  temperature  of  the  steam  to  be 
350°,  and  that  of  the  feed  water  110°.  How  many  pounds  of  water 
would  a  pound  of  the  coal  evaporate  under  those  conditions  ? 

We  have:  Hw  =1091.7  +  0.305(350  —  32)  —  (110  —  32)  = 
1110.69   units   required   per   pound   of   water   evaporated;    hence 
11,974 
1110139  =  l^'^  pounds  of  water  evaporated  per  pound  of  coal. 


30  THE  ELEMENTS  OF  STEAM  ENGINEERING 

45.  At  their  meeting  in  1899,  The  American  Society  of  Mechani- 
cal Engineers  revised  the  code  for  conducting  steam  boiler  trials. 

By  this  code  the  "  unit  of  evaporation  "  is  stated  to  be,  "  one 
pound  of  water  at  212°  F.  evaporated  into  dry  steam  at  the  same 
temperature."  This  unit  is  equivalent  to  965.7  B.  T.  IT. 

The  code  recommends  that  the  capacity  of  a  boiler  be  expressed 
in  terms  of  the  "number  of  pounds  of  water  evaporated  per  hour 
from  and  at  212°."  It  does  not,  however,  abandon  the  widely 
recognized  measure  of  capacity  of  stationary  boilers  expressed  in 
terms  of  "  boiler  horse-power." 

46.  The  unit  of  commercial  horse-power  developed  by  a  boiler  was 
adopted  by  a  Commission  of  the  Centennial  Exhibition  in  1876, 
and  is  taken  to  be  "  an  evaporation  of  30  pounds  of  water  per  hour 
from  a  feed  water  temperature  of  100°  F.  into  steam  at  70  pounds 
pressure  per  gauge." 

47.  It  is  the  common  practice  to  reduce  the  results  of  boiler  tests 
to  the  common  standard  of  weight  of  water  evaporated  by  the  unit 
weight  of  the  combustible  portion  of  the  fuel,  the  evaporation  being 
considered  to  have  taken  place  at  mean  atmospheric  pressure,  and 
at  the  temperature  due  that  pressure,  the  feed  water  being  also 
assumed  to  have  been  supplied  at  that  temperature.   This  is,  in  tech- 
nical language,  said  to  be  the  equivalent  evaporation  from  and  at 
atmospheric  pressure,  or  "from  and  at  212°  F." 

48.  In  determining  the  commercial  rating,  or  horse-power  of 
boilers  from  tests,  the  evaporative  results  obtained  under  the  con- 
ditions of  the  trials  must  be  brought  to  the  uniform  standard  of 
comparison  of  "from  and  at  212°  F."     This  can  readily  be  done 
by  means  of  a  "  factor  of  evaporation  "  which  is  immediately  de- 
ducible  from  the  results  of  any  trial  test. 

In  the  example  last  given  the  steam  was  generated  from  feed 
water  of  110°  temperature  into  steam  of  350°,  and  the  units  re- 
quired for  each  pound  of  water  evaporated,  or  II  w,  equalled  1110.69. 
It  has  been  shown  that  if  feed  water  at  212°  be  evaporated  into 
steam  at  212°  there  are  required  only  965.7  units,  therefore  the 
conditions  of  the  example  would  produce  an  evaporating  result 

r  =1.15  times  greater  from  and  at  212°  ;   or  the  equivalent 


evaporation    from    and    at    212°    per   pound    of    coal    would    be 
10.78  X  1.15  =  12.4  pounds. 


FUELS  AND  COMBUSTION  31 

TT 

In  any  case,  then,  the  factor  of  evaporation  is  equal  to  gg^ ; 

and  the  evaporative  results  obtained  by  the  tests  of  the  boiler,  mul- 
tiplied by  this  factor,  gives  the  equivalent  evaporation  from  and 
at  212°. 

49.  The  unit  of  commercial  boiler  horse-power  has  been  stated 
to  be  the  evaporation  of  30  pounds  of  water  per  hour  from  a  feed 
water  temperature  of  100°  F.  into  steam  at  70  pounds  pressure  per 
gauge,  or  an  absolute  pressure  of  84.7  pounds.     The  temperature 
of  steam  at  84.7  pounds  is  found  by  calculation  to  be  316°.     We 
have,  under  these  conditions, 

Hw  —  1091.7  +  0.305(316  —  32)  —  (100  —  32)  .==  1110.32  units. 

Then  the  factor  of  evaporation  =  ^^  =  1^^2  =  1.1498. 

Therefore,  an  evaporation  of  30  pounds  water  from  100°  into  steam 
at  316°  is  equivalent  to  the  evaporation  of  30  X  1.1498  =  34.5 
pounds  from  and  at  212°. 

50.  The  practical  method  of  determining  the  commercial  horse- 
power of  a  boiler  is  to  ascertain  by  test  the  number  of  pounds  of 
water  evaporated  per  hour  under  the  working  conditions,  and  then, 
by  means  of  the  factor  of  evaporation,  find  the  equivalent  number  of 
pounds  from  and  at  212°.     This  latter  quantity  divided  by  34.5 
will  give  the  horse-power  of  the  boiler. 

PKOBLEMS. 

8.  The  average  composition  of  air  by  weight  is  77  parts  of  N" 
and  23  parts  of  0.     How  many  pounds  of  air  are  theoretically 
necessary  for  the  complete  combustion  of  one  pound  of  carbon? 

Ans.  11.6  Ibs. 

9.  Find  the  number  of  pounds  of  air  necessary  for  the  complete 
combustion  of  one  pound  of  hydrogen.  Ans.  35  Ibs. 

10.  Write  the  formula  for  the  total  heat  of  combustion  of  a  pound 
of  fuel,  and  explain  its  derivation. 

11.  Coal  containing  92  per  cent  of  C,  4  per  cent  of  H,  3  per  cent 
of  0,  and  1  per  cent  of  H20,  is  burned  in  a  furnace  with  a  supply  of 
air  of  22.5  pounds  per  pound  of  coal.     Temperature  of  the  entering 
air  50°,  temperature  of  discharged  gases  600°,  and  temperature  of 
the  feed  water  112°.     The  specific  heats  of  N,  C02,  superheated 


32  THE  ELEMENTS  OF  STEAM  ENGINEERING 

steam,  and  0,  are  0.244,  0.217,  0.475,  and  0.218,  respectively.  Find : 
(a)  Temperature  of  furnace.  (&)  Heat  available  for  generating 
steam  per  pound  of  coal,  (c)  The  number  of  pounds  of  coal  neces- 
sary to  evaporate  50  pounds  of  water  into  steam  of  270°  tempera- 
ture. Ans.  (a)  2810°.  (&)  12,474  units,  (c)  4.35  pounds. 

12.  An  analysis  of  coal  shows  its  composition  to  be  91.5  per  cent 
C,  3.5  per  cent  H,  and  2.6  per  cent  0.     How  much  air  will  be  theo- 
retically necessary  per  pound  to  burn  it  ?       Ans.  11.714  pounds. 

13.  (a)   How  many  pounds  of  water  will  a  pound  of  the  coal  of 
Example  12  evaporate  from  feed  water  of  120°  temperature  into 
steam  of  373°  temperature?     (b)  What  is  the  evaporative  power  of 
this  coal  from  and  at  212°  ? 

Ans.   (a)   13.75  pounds.     (6)   15.80  pounds. 

14.  Assuming  the  whole  of  the  14,500  thermal  units  of  a  pound 
of  carbon  were  available  for  useful  work,  how  many  pounds  per 
I.  H.  P.  per  hour  would  be  required?  Ans.  0.176  pound. 


CHAPTER  IV. 
EFFICIENCY. 

51.  The  total  work  performed  by  any  machine  is  not  usefully 
employed,  as  there  are  always  causes  which  occasion  a  waste  of 
work,  notably  that  expended  in  overcoming  the  friction  of  the 
machine. 

The  ratio  of  the  useful  work  performed  to  the  total  energy  ex- 
pended is  called  the  efficiency  of  the  machine,  or 

Useful  work  performed 
~~  Total  energy  expended 

52.  Before  any  further  consideration  of  the  question  of  efficiency, 
a  brief  reference  will  be  made  to  Carnot's  Reversible  Cycle  or  Per- 
fect Heat  Engine. 

Any  series  of  operations  which  returns  upon  itself  is  termed  a 
cycle,  and  in  physics  a  cycle  is  a  series  of  operations  by  which  a 
substance  is  brought  back  to  its  initial  state.  If  a  gas  in  a  working 
process  passes  through  a  series  of  heat  changes  and  finally  returns 
to  its  initial  condition,  the  changes  constitute  a  cycle. 

The  perfect  heat  engine  of  Carnot  is  imaginary  in  conception,  as 
it  is  supposed  to  be  built  of  materials  which  have  no  practical  ex- 
istence, such  as  are  perfectly  conducting  and  perfectly  non-conduct- 
ing. The  study  of  the  results  of  the  perfect  heat  engine  is  of  great 
importance,  however,  from  the  fact  that  it  enables  us  to  point  out 
the  imperfections  of  existing  engines,  and  to  assign  to  each  its 
share  of  the  responsibility  for  the  waste  of  heat. 

A  full  discussion  of  the  operations  of  the  Carnot  cycle  may  be 
found  in  any  treatise  on  the  thermodynamics  of  the  steam  engine, 
and  only  a  reference  incidental  to  an  understanding  of  engine  effi- 
ciency is  here  attempted. 

The  gas  assumed  to  be  used  in  the  Carnot  cycle  is  air,  and  during 
the  changes  of  state  which  it  undergoes  heat  is  expended  in  the 
performance  of  external  work  of  raising  a  piston,  and  on  the  inter- 
nal work  of  changing  the  temperature  or  the  molecular  condition 
of  the  gas.  During  the  process,  by  which  the  gas  is  returned  to  its 
initial  state,  work  is  performed  upon  the  gas  and  the  heat  expended 
3 


34  THE  ELEMENTS  OF  STEAM  ENGINEERING 

in  the  internal  work  will  have  been  returned  or  rejected.  The  final 
result  is  the  performance  of  external  work  equal  to  the  total  heat 
expended,  diminished  by  the  heat  of  the  internal  work. 

It  will  be  shown  later  on  that  no  engine  can  be  devised  that  will 

rp    rp 

show  a  greater  efficiency  than  - J-™ — 2 ,  in  which  T±  is  the  absolute 

•*  i 

temperature  of  the  source  of  heat,  and  T2  the  absolute  temperature 
of  the  source  of  cold.  The  engine  producing  this  efficiency  is  known 
as  the  ideal  engine. 

T  —  T 

The  expression  —^ —  l  for  the  efficiency  would  have  its  maxi- 
mum value  of  unity  when  T2  =  0,  or  when  the  source  of  cold  has 
the  absolute  zero  of  temperature,  a  result  practically  unattainable. 
The  best  we  can  do  is  to  make  the  efficiency  of  the  engine  approach 
unity  by  making  T^  —  T2  as  nearly  as  possible  qual  to  T19  which  is 
practically  accomplished  by  making  2\  as  large  and  T2  as  small  as 
possible.  Unfortunately,  the  practical  range  of  temperature  in  the 
cylinder  of  a  steam  engine  is  very  limited  and  the  efficiency,  there- 
fore, very  low.  The  maximum  temperature  which  could  possibly 
be  available  is  that  of  the  products  of  combustion  of  the  fuel,  and 
if  there  were  practical  means  of  utilizing  the  temperature  of  furnace 
gases,  some  material  other  than  iron  would  have  to  be  used  in  the 
construction  of  engines.  We  are,  therefore,  limited  for  the  maxi- 
mum temperature  to  that  of  the  steam  in  the  boiler,  and  for  the 
minimum,  a  condenser  temperature  of  about  126°  F.,  corresponding 
to  an  absolute  pressure  of  2  pounds. 

The  object  in  a  steam  engine  is  to  obtain  external  work  by  an 
expenditure  of  heat;  and  while  the  ideal  efficiency  cannot  be  ob- 
tained in  practice,  yet  it  is  the  aim  of  engineers  to  approach  it  as 
nearly  as  possible. 

53,  Losses  Affecting  Efficiency. — The  principal  causes  which  re- 
duce the  efficiency  of  the  steam  engine  below  that  of  the  ideal  engine 
may  be  enumerated  as  follows : 

1.  Steam  is  far  from  being  a  perfect  gas,  and  its  employment  as 
the  medium  is  primarily  disadvantageous.     It  is  impossible  to  com- 
press the  condensed  steam  to  its  initial  state,  necessitating  a  reheat- 
ing in  the  boiler. 

2.  Steam  is  not  supplied  to  the  cylinder  at  the  temperature  of  the 
furnace,  the  source  of  heat. 


EFFICIENCY  35 

3.  Steam  is  not  rejected  to  the  condenser,  the  source  of  cold,  at 
the  condenser  temperature  and  pressure,  but  suffers  a  fall  in  both 
particulars,  in  consequence  of  incomplete  expansion. 

4.  Owing  to   inevitable   leakages   an   equivalence   in   water   to 
the  steam  supplied  to  the  cylinder  is  not  returned  to  the  boiler, 
necessitating  an  addition  to  the  feed  of  water  of  much  lower  tem- 
perature which  must  be  raised  to  the  temperature  of  the  boiler 
steam. 

5.  Initial  condensation  of  steam  in  the  cylinder  occasions  a  loss, 
though  this  is  partially  recompensed  by  re-evaporation  during  ex- 
pansion. 

6.  The  unavoidable,  yet  practically  desirable,  clearance  space  in 
the  cylinder  must  be  filled  with  steam  at  every  stroke,  and  this  steam 
does  no  work  except  during  the  period  of  expansion. 

7.  "  Priming  "  in  the  boiler  occasions  an  admixture  of  steam  and 
water,  by  which  particles  of  water  are  carried  into  the  cylinder  and 
thence  to  the  condenser  without  doing  any  work,  and  abstracts  heat 
from  the  cylinder  in  their  efforts  to  evaporate. 

8.  Heat  is  lost  by  radiation,  and  by  wire  drawing  of  the  steam. 

9.  Free  and  unrestricted  expansion  of  the  steam,  resulting  in  the 
performance  of  no  external  work,  occasions  a  loss,  as  instanced  by 
the  "  drop  "  from  cylinder  to  cylinder  in  multiple-expansion  engines. 

10.  Loss  due  to  friction  of  the  working  parts  and  of  friction  of 
the  steam  in  pipes. 

54.  Condensation  and  Re-evaporation  of  Steam  in  Cylinders  and 
Methods  of  Reducing  Losses  Therefrom. — When,  at  the  beginning 
of  the  stroke,  fresh  boiler  steam  enters  the  cylinder,  which,  has  just 
been  cooled  by  exhaust  connection  with  the  condenser,  initial  con- 
densation ensues  which  is  not  shown  by  the  indicator  diagram.  If 
the  cut-off  is  earlier  than  half  stroke  it  is  probable  that  a  further 
condensation,  due  to  the  work  done  and  the  difference  in  tempera- 
tures of  the  steam  and  cylinder,  occurs  during  the  earlier  part  of 
the  expansion,  resulting  in  the  lowering  of  the  expansion  curve.  As 
the  stroke  proceeds,  the  latent  heat  of  the  liquefied  steam  raises  the 
temperature  of  the  cylinder  walls,  and  the  reduction  of  pressure 
due  to  the  expansion  lowers  the  temperature  of  the  steam,  with  the 
result  that  abstraction  of  heat  from  the  cylinder  ensues,  causing  a 


36  THE  ELEMENTS  OF  STEAM  ENGINEERING 

partial  re-evaporation  of  the  water  and  the  consequent  raising  of 
the  expansion  curve.  This  re-evaporation  does  not  ever  make  up 
for  the  loss  due  to  condensation,  and  there  is,  therefore,  always  a 
quantity  of  water  rejected  at  release,  and  owing  to  the  low  pressure 
during  exhaust  much  of  it  is  re-evaporated,  resulting  in  the  crea- 
tion of  back  pressure. 

There  are  four  methods  of  reducing  the  losses  from  cylinder  con- 
densation and  re-evaporation,  viz. :  1.  The  use  of  superheated  steam, 
2.  Steam  jacketing  the  cylinder.  3.  Quick  running.  4.  Com- 
pounding, or  allowing  the  steam  to  expand  in  two  or  more  cylinders, 
thus  reducing  the  range  of  temperature  in  each. 

55.  Superheated  Steam. — Steam  in  the  condition  in  which  it  is 
ordinarily  produced  and  used  is  said  to  be  saturated,,  the  name  im- 
plying that  it  is  a  saturated  vapor  having  the  maximum  density — 
and  hence  the  smallest  volume  per  pound — consistent  with  its  pres- 
sure or  with  its  temperature;  it  is  steam  at  the  point  of  condensa- 
tion, and  any  reduction  in  the  temperature  will  cause  liquefaction. 
If  saturated  steam  be  heated  so  that  its  temperature  rises  above  that 
of  the  saturation  point  it  is  said  to  be  superheated,  and  the  employ- 
ment of  such  steam  in  cylinders  is  the  most  effective  means  of  reduc- 
ing the  loss  from  liquefaction.  The  behavior  of  superheated  steam 
approaches  much  nearer  that  of  a  perfect  gas  than  does  that  of 
saturated  steam,  and  having  a  specific  heat  at  constant  pressure  of 
0.48,  it  requires  less  than  half  a  thermal  unit  to  raise  the  tempera- 
ture of  a  pound  to  the  extent  of  one  degree. 

It  must  be  understood  that  the  addition  of  heat  to  saturated  steam 
will  raise  the  temperature  and  leave  the  pressure  constant,  provided 
the  steam  be  allowed  to  expand  as  the  heat  is  added.  The  practical 
method  of  superheating  steam  is  to  cause  it  to  pass,  on  its  way  from 
boiler  to  cylinder,  through  coils  surrounded  by  hot  gases  from  the 
furnace.  The  steam  absorbs  heat  from  the  gases  with  a  consequent 
rise  in  temperature,  the  pressure  remaining  constant  from  the  fact 
that  the  steam  is  used  almost  as  fast  as  it  is  generated,  and  the  dis- 
placement of  the  piston  in  the  cylinder  causes  a  virtual  extension 
of  the  volume  of  the  boiler,  thus  allowing  the  steam  to  expand  and 
the  pressure  to  remain  constant. 

The  total  heat  of  superheated  steam  is  equal  to  the  total  heat  of 
the  steam  in  the  saturated  condition,  plus  the  quantity  of  heat 
0.48 (ts  —  t)  required  to  raise  the  temperature  of  the  steam  from 


EFFICIENCY  37 

that  due  its  saturated  condition  to  the  temperature  ts  of  its  super- 
heat, the  pressure  remaining  constant;   or, 

Ha  =  HT+QAS(t8  —  t)  =  1091.7  +  0.305(2— 32)  -\-QAS(ts—t). 

Steam  entering  the  cylinder  in  a  superheated  condition  is  able  to 
give  up  to  the  cylinder  walls  the  whole  of  its  superheat  without 
undergoing  any  liquefaction,  and  if  the  superheat  be  sufficiently 
high  the  loss  from  liquefaction  may  be  entirely  prevented. 

The  amount  of  superheat  necessary  to  prevent  liquefaction  in  any 
particular  case  is  not  susceptible  of  exact  calculation,  but  experi- 
ments made  by  Mr.  Ripper  on  the  behavior  of  superheated  steam  in 
a  small  engine  indicated  that  for  each  1  per  cent  of  wetness  at 
cut-off  7.5°  F.  of  superheat  must  be  present  in  the  steam  on  admis- 
sion to  the  cylinder  in  order  to  render  the  steam  dry  at  cut-off. 

For  example,  if  a  small  simple  engine  using  saturated  steam  at 
90  Ibs.  pressure  shows  a  condensation  of  20  per  cent  of  the  steam 
up  to  cut-off,  there  would  be  required  7.5°  X  20  =  150°  of  super- 
heat to  secure  dry  steam  at  cut-off.  As  the  temperature  of  saturated 
steam  of  90  Ibs.  pressure  is  320°,  the  temperature  of  the  super- 
heated steam  supplied  to  the  cylinder  would  have  to  be  320  + 
150  =  470°. 

The  loss  from  initial  condensation  in  small  engines  is  greater  than 
in  large  ones,  for  the  reason  that  the  smaller  the  diameter  of  the 
cylinder  the  greater  the  ratio  of  the  interior  surface  to  the  volume. 
That  is,  the  smaller  the  cylinder  the  greater  the  surface  to  be  alter- 
nately heated  and  cooled  per  pound  of  steam  used,  and  hence  the 
greater  the  liquefaction. 

56.  Steam  Jacketing.— The  jacket  of  a  cylinder  is  usually  formed 
by  an  annular  space  between  the  working  barrel  of  the  cylinder  and 
the  cylinder  proper.  Steam  of  boiler  pressure  is  circulated  through 
the  jacket,  the  object  being  to  supply  heat  to  the  interior  surface  of 
the  cylinder,  and  thus  reduce  condensation  and  promote  re-evapora- 
tion. This  transfer  of  heat  from  the  jacket  to  the  walls  of  the 
cylinder  causes  a  certain  amount  of  the  jacket  steam  to  liquefy,  but 
this  is  drained  and  returned  to  the  boiler  without  material  waste. 
The  virtue  of  the  jacket  lies  in  the  fact  that  the  latent  heat  due  to 
the  condensation  of  its  steam  is  delivered  to  the  steam  within  the 
cylinder,  with  the  result  that  the  working  steam  is  kept  compara- 
tively dry,  and  being  a  poor  conductor  and  radiator  of  heat  when  in 
that  condition,  the  cylinder  walls  are  unable  to  give  heat  to  and 


38  THE  ELEMENTS  OF  STEAM  ENGINEERING 

receive  it  from  the  steam,  and  are  thereby  deprived  of  the  power  of 
taking  up  the  extremes  of  temperature.  The  greatest  benefit  to  be 
derived  from  jacketing  is  with  slow  speed  simple  engines,  its  value 
being  less  with  multiple  expansion  engines,  and  almost  valueless 
with  engines  of  high  rotational  speeds. 

57.  Quick  Running. — It  is  readily  seen  that  with  engines  making 
a  large  nurnber  of  revolutions  per  minute,  there  is  insufficient  time 
to  make  the  transfer  of  heat  to  and  from  the  cylinder  walls,  and 
condensation  is  therefore  diminished. 

58.  Compounding. — The  introduction  of  higher  steam  pressures, 
with  increased  rates  of  expansion,  produced  such  wide  ranges  of 
temperature  in  a  single  cylinder  that  the  remedies  already  consid- 
ered for  cylinder  condensation  failed  in  their  application.     A  di- 
vision of  this  wide  range  of  temperature  was  then  resorted  to  by 
allowing  the  steam  to  expand  successively  through  two,  three,  and 
sometimes  four  cylinders,  thus  introducing  the  double,  the  triple, 
and  the  quadruple  expansion  engines. 

For  example,  the  range  of  temperature  in  a  triple  expansion 
engine  using  steam  of  initial  pressure  of  160  Ibs.  would  be  from 
363°  to  about  152°,  corresponding  to  a  back  pressure  in  the  con- 
denser of  4  Ibs. ;  and  a  ratio  of  expansion  of  13  could  easily  be 
attained.  If  such  an  expansion  were  attempted  in  a  single  cylinder 
the  fall  in  temperature  would  be  363°  — 152°  =  211°,  a  range 
which  would  cause  excessive  condensation,  besides  other  evils  which 
will  not  be  noticed  now.  With  the  triple  expansion  engine  this 
range  of  temperature  could  be  about  equally  divided  between  the 
three  cylinders,  and  the  initial  condensation  much  reduced. 

59.  Efficiencies  of  Engines  and  Boilers. — There  are  a  number  of 
ways  of  expressing  the  efficiencies  of  engines  and  boilers,  and  as 
such  expressions  are  the  means  of  estimating  and  comparing  the 
economies  of  engine  and  boiler  systems,  they  should  be  carefully 
studied  in  order  that  there  should  be  no  confusion  in  their  appli- 
cation. 

60.  Pounds  of  Coal  per  I.  H.  P.  per  Hour. — This  is  a  very  com- 
mon standard  of  economy  of  an  engine  and  boiler  when  considered 
as  one  machine.     Its  value  may  be  averaged  as  3  for  non-condensing 
engines,  2.25  to  2.5  for  condensing  engines,  and  1.5  to  2  in  multiple- 
expansion  engines. 


EFFICIENCY  39 

61.  Pounds  of  Steam  per  I,  H.  P.  per  Hour. — This  standard  is  a 
measure  of  the  performance  of  an  engine,  independently  of  the 
boiler.     The  value  of  this  standard  varies  from   12.5  to   16   for 
multiple-expansion  engines,  and  from  24  to  30  for  non-condensing 
engines. 

62.  Thermal  Units  per  I.  H.  P.  per  Minute. — The  number  of  ther- 
mal units  expended  per  I.  H.  P.  per  minute  is  a  standard  of  engine 
performance  much  used  in  preference  to  the  standard  of  weight  of 
steam  per  I.  H.  P.  per  hour. 

63.  Thermal  Efficiency. — Knowing  the  number  of  thermal  units 
per  I.  H.  P.  per  minute,  the  standard  of  thermal  efficiency  may  be 
expressed  as  of  the  engine  alone,  or  it  may  be  made  to  include  both 
boiler  and  engine. 

The  development  of  one  horse-power  is  equivalent  to  the  expen- 

33  000 
diture  of      '  0     =  42.42  thermal  units,  so  that  the 

I  /  O 

Thermal  Efficiency  of  Engine 


Thermal  Efficiency  of  Eng.  and  Boiler  = 


Thermal  units  in  steam  per  I.  H.  P.  per  min. 
42.42 


Thermal  units  from  coal  per  I.H.  P.  per  min. 

As  an  example  of  the  first  efficiency  take  the  case  of  a  non-con- 
densing engine  working  between  the  temperatures  of  324°  and  220°, 
corresponding  to  an  initial  pressure  of  95  Ibs.  and  a  back  pressure 
of  17  Ibs.,  and  using  24  Ibs.  steam  per  I.  H.  P.  per  hour.  If  the 
feed  water  were  taken  at  a  temperature  of  32°,  the  heat  expended 
per  pound  of  steam  would  be  the  total  heat  of  formation  of  steam 
at  324°,  but  if  the  feed  water  be  heated  to  the  temperature  of  the 
exhaust,  we  must  subtract  220  —  32  =  188  units  from  that  amount; 
hence : 

4.9  A.O  v   fift 

Thermal  Efficiency  of  Engine  =  —  =  0.107. 

24  [1091 .7  +  0.305(324  -  32)  -  (220  -  32)  J 

If  the  example  be  that  of  a  triple-expansion  engine  working  be- 
tween the  temperatures  of  366°  and  126°,  corresponding  to  an  initial 
pressure  of  165  Ibs.,  and  a  condenser  pressure  of  2  Ibs. ;  and  if  the 
consumption  of  steam  be  12.5  Ibs.  per  I.  H.  P.  per  hour,  we  will 
have: 

Thermal  Efficiency  of  Engine  =  —  =  0.195. 

12.5[1091.7  +  0.305(366  -  32)  -  (126  -  32)] 

It  is  seen  from  this  that  10.7  per  cent  and  19.5  per  cent  are  good 
values  for  the  efficiency  of  the  engine  alone. 


40  THE  ELEMENTS  OF  STEAM  ENGINEERING 

As  an  example  of  the  combined  engine  and  boiler  efficiency,  we 
will  suppose  the  expenditure  of  coal  to  be  2.4  Ibs.  per  I.  H.  P.  per 
hour,  and  the  thermal  value  of  the  coal  to  be  14,200  units  per  pound. 
Then, 

Thermal  Efficiency  of  Engine  and  Boiler  =^'^  *  ^  =  0.0747. 

14:j^UU    /\     /C.4- 

Good  values  for  the  thermal  efficiency  of  the  engine  and  boiler 
combined  are  from  7  per  cent  to  10  per  cent. 

64.  Relative  Engine  Efficiency. — If  the  thermal  efficiency  be 
compared  with  the  efficiency  of  the  ideal  engine  of  the  same  range 
of  temperature,  the  result  is  the  standard  of  relative  efficiency. 

The  ideal  efficiency  of  the  triple-expansion  engine  above  con- 
sidered is 

T,—  T2  _  (366  +  461)  — (126  +  461)  _  0  9q 
T,  366  +  461 

0  195 
and  the  relative  efficiency  is    Q  ^    —  0.673,  or  67.3  per  cent  of  the 

ideal  efficiency. 

65.  Mechanical  Efficiency  of  an  Engine. — Brake  horse-power  is 
the  power  developed  by  an  engine  independently  of  the  power  ab- 
sorbed in  friction  in  driving  the  engine  itself.     Indicated  horse- 
power includes  this  frictional  work;  hence, 

Brake  Horse-power 
Mechanical  Efficiency  ==  Indicated  Horse-power 

and  its  value  should  vary  from  85  per  cent  to  90  per  cent. 

66.  Boiler  Efficiency. — A  standard  of  boiler  efficiency  in  general 
use  is  that  of  the  number  of  pounds  of  water  it  evaporates  per 
pound  of  coal,  corresponding  to  the  standard  of  engine  efficiency 
expressed  by  the  number  of  pounds  of  steam  it  consumes  in  the  de- 
velopment of  an  indicated  horse-power. 

67.  The  real  efficiency  of  a  boiler  is  the  product  of  two  efficien- 
cies, that  of  the  heating  surface  and  that  of  the  combustion. 

a      ,  Heat  absorbed  by  water  per  Ib.  of  fuel 

Efficiency  of  Heating  Surface  =  —  — . 

Heat  available  per  Ib.  of  fuel 

Efficiency  of  Combustion  =   Heat  available  per  Ib.  of  fuel    ^  therefore> 
Heat  contained  in  a  Ib.  of  fuel 

Boiler  Efficiency  =  Heating  Surface  Efficiency  x  Combustion  Efficiency 
_  Heat   absorbed   by  water  per  Ib.  of  coal 
Heat  contained  in  a  Ib.  of  coal 


EFFICIENCY  41 

This  may  be  better  understood  from  a  numerical  example. 

EXAMPLE  I.  —  A  boiler  using  coal  containing  90  per  cent  C,  3  per 
cent  H,  and  6  per  cent  0,  evaporates  10  Ibs.  of  water  per  pound  of 
coal  into  steam  at  341°  from  feed  at  122°.  3130  units  are  wasted 
through  the  smoke  pipe  per  pound  of  coal  used.  It  is  required  to 
find:  (a)  The  efficiency  of  the  heating  surface.  (&)  The  efficiency 
of  the  combustion,  (c)  The  efficiency  of  the  boiler. 

Solution: 

h  =  14,500  [0.9  -f  4.28  (  0.03  -  ^  \~\  =  14,446  units  con- 

tained in  a  pound  of  the  fuel. 
Hw—  1091.7  +  0.305(341  —  32)  —  (122  —  32)  =1096  units 

required  to  evaporate  one  pound  of  water. 

1096  X  10  =  10,960  units  absorbed  by  the  water  per  pound  of  coal. 
14,446  —  3130  =  11,316  units  available  per  pound  of  coal. 

Efficiency  of  heating  surface  =  =  0.9686. 


Efficiency  of  combustion  =  jg    =  0.7835. 

Efficiency  of  boiler  =  0.9686  X  0.7835  =  0.7588. 

Or,  we  might  have  proceeded  as  follows  : 
Heat  available  per  pound  of  coal  =  1^316  =  Ia8g  ]bg 

MW 

should  have  been  evaporated  per  pound  of  coal  from  the  available 
heat. 

Efficiency  of  heating  surface  =  T-QQ  =  0.9686. 


Heat  units  in  pound  of  coal  =  146 


be  evaporated  if  all  the  heat  in  the  coal  were  utilized. 

10  33 

Efficiency  of  combustion  =  -..,'      =  0.7835. 

68.  Total  Efficiency  of  System.  —  The  percentage  of  indicated  work 
performed  by  the  engine  from  a  given  weight  of  coal  is  expressed  by 
the  product, 

Efficiency  of  Boiler  X  Thermal  Efficiency  of  Engine, 

and  the  percentage  of  the  work  performed  that  is  applied  to  the 
shaft  is  : 

Boiler  Efficiency  x  Thermal  Engine  Efficiency  x  Mechanical  Efficiency. 


42  THE  ELEMENTS  OF  STEAM  ENGINEERING 

69.  To  illustrate  the  imperfection  of  the  steam  engine  we  will 
consider  its  efficiency  by  means  of  a  numerical  example  from  the 
conditions  of  actual  practice. 

EXAMPLE  II.  —  A  high-speed  non-condensing  engine  works  with 
an  initial  absolute  pressure  of  95  Ibs.,  the  temperature  of  which  is 
324°.  The  feed  water  has  a  temperature  of  180°.  By  a  measurement 
of  the  feed  water  it  is  found  that  25  Ibs.  of  steam  are  used  per 
I.  H.  P.,  and  a  test  of  the  boiler  shows  that  10  Ibs.  of  water  are 
evaporated  per  pound  of  coal.  The  thermal  value  of  the  coal  used  is 
14,320  units  per  pound.  It  is  required  to  find  the  efficiency  of  the 
system. 

The  heat  required  for  the  production  of  one  pound  of  the  steam  is 
Hw  =  1091.7  +  0.305(324  —  32)  —  (180  —  32)  =  1032.76  units, 

therefore  one  pound  of  the  coal  should  evaporate  i  QQO  75  —  13.88 
pounds  of  water. 

Efficiency  of  Boiler  =  |J£j  —  0.72. 

lO.oo 

1032  76  ^  25 

^-™  —    -  =  430.3  thermal  units  required  per  I.  H.  P.  per 

minute. 
The  development  of  one  horse-power  is  equivalent  to  the  expendi- 

qq  ooo 

ture  of  ~£r  =  4^.42  thermal  units;  hence: 


Thermal  efficiency  of  engine  =  ^-^  =  0.098. 

The  efficiency  of  the  boiler  is  0.72,  and  as  only  9.8  per  cent  of  the 
heat  supplied  to  the  engine  is  converted  into  work,  the  efficiency  of 
the  system  is  0.098  X  0.72  =  0.07  ;  or  only  7  per  cent  of  the  heat  of 
the  fuel  is  converted  into  mechanical  work;  and  if  the  mechanical 
efficiency  of  the  engine  is  87  per  cent,  we  would  have  only 
0.07  X  0.87  X  100  =  6.09  per  cent  of  the  heat  energy  of  the  fuel 
applied  to  the  shaft. 

PEOBLEMS. 

15.  It  is  claimed  that  the  best  modern  engines  develop  one 
I.  H.  P.  by  the  expenditure  of  1.6  Ibs.  of  coal.  Comparing  this 
result  with  that  obtained  from  carbon  in  problem  14,  determine  the 
efficiency  of  the  modern  engine.  Arts.  11  per  cent. 


EFFICIENCY  43 

16.  Diameter  of  cylinder  24",  stroke  30",  cut-off  f  stroke,  revo- 
lutions per  minute  125.     Thermal  value  of  the  fuel  15,000  units 
per  pound,  of  which  4000  are  wasted.     Temperature  of  feed  water 
102°,  and  of  steam  327.5°,  corresponding  to  an  absolute  pressure 
of  100  Ibs.     Eelative  volumes  of  steam  and  water  at  100  Ibs.  pres- 
sure are  271  :  1.     Eequired:    (a)    The  pounds  of  coal  used  per 
hour,     (b)  The  evaporative* power  of  the  coal  from  and  at  212°. 

Ans.   (a)   1030.26  Ibs.     (b)    11.39  Ibs. 

17.  The  efficiency  of  a  given  boiler  is  70  per  cent,  and  .of  the 
engine  10  per  cent.     The  composition  of  the  fuel  used  is  90  per 
cent  C,  6  per  cent  H,  and  4  per  cent  0.     How  much  coal  must  be 
burned  per  hour  to  develop  1000  I.  H.  P.  ?  Ans.  2208  Ibs. 

18.  A  boiler  is  to  evaporate  45  cu.  ft.  of  water  per  hour  into 
steam  of  300°  temperature  from  feed  water  of  100°.     Average  rate 
of  transmission  per  sq.  ft.  of  heating  surface  per  hour  is  3000  units. 
Efficiency  of  boiler  70  per  cent,  and  thermal  value  of  one  pound  of 
the  coal  used  is   15,000  units.     Eequired:    (a)    Heating  surface. 
(b)   The  evaporative  power  of  the  coal  from  and  at  212°. 

Ans.   (a)   1036.55  sq.  ft.     (6)   10.87  Ibs. 

19.  A  boiler  using  coal  containing  88  per  cent  C,  5  per  cent  H, 
and  4  per  cent  0,  evaporates  9.5  Ibs.  of  water  per  pound  of  coal  into 
steam  of  366°  from  feed  water  of  152°.     4550  units  are  wasted  per 
pound  of  coal  used.     Find:  (a)  Efficiency  of  heating  surface,    (b) 
Efficiency  of  combustion,     (c)  Efficiency  of  the  boiler. 

Ans.   (a)  92.75  per  cent,    (b)  70.73  per  cent,    (c)  65.60  per  cent. 

20.  A  non-condensing  engine  working  with  an  initial  absolute 
steam  pressure  of  100  Ibs.,  the  temperature  of  which  is  328°,  uses 
26  Ibs.  steam  per  I.  H.  P.  per  hour.     The  boiler  evaporates  9.5  Ibs. 
of  water  per  pound  of  coal,  the  temperature  of  the  feed  being  202°. 
The  thermal  value  of  the  coal  used  is  14,400  units  per  pound,  and 
it  is  found  that  6  per  cent  of  the  heat  energy  of  the  fuel  is  applied 
to  the  shaft.     Find:   (a)   Efficiency  of  the  boiler,     (b)   Thermal 
efficiency  of  the  engine,     (c)   Mechanical  efficiency  of  the  engine. 

Ans.   (a)  66.76  per  cent,    (b)  9.67  per  cent,    (c)  93.00  per  cent. 

21.  An  engine  and  boiler  test  shows  a  heating  surface  efficiency 
of  96  per  cent.     There  are  11,400  units  available  for  generating 
steam,  and  3100  units  wasted,  per  pound  of  coal.     The  tempera- 


44  THE  ELEMENTS  OF  STEAM  ENGINEERING 

ture  of  the  steam  was  342°,  and  of  the  feed  water  122°.  .  Find: 
(a)  Efficiency  of  combustion.     (&)  Efficiency  of  boiler,     (c)  The 
number  of  pounds  of  water  evaporated  per  pound  of  coal. 
Ans.   (a)  78.62  per  cent.    (6)  75.48  per  cent,    (c)  9.98  per  cent. 

22.  A   triple-expansion   engine   with   forced   draft   develops   20 
I.  H.  P.  per  sq.  ft.  of  grate  surface,  burns  35  Ibs.  of  coal  per  sq.  ft. 
of  grate, per  hour,  and  consumes  14  Ibs.  of  steam  per  I.  H.  P.  per 
hour.     Absolute  boiler  pressure  160  Ibs.,  the  temperature  of  which 
is  363°.     Temperature  of  feed  water  150°,  and  12,010  units  are 
available  for  generating  steam  per  pound  of  coal.    Find :  (a)  The 
efficiency  of  the  heating  surface.    (&)  The  evaporative  power  of  the 
boiler  from  and  at  212°.  Ans.   (a)  0.716.    (6)  9.94 Ibs. 

23.  A  boiler  generates  steam  of  temperature  of  366°  from  feed 
water  at  152°.     There  are  11,846  units  available  for  generating 
steam  per  pound  of  coal.     The  efficiency  of  the  heating  surface  is 
88  per  cent-,  and  the  consumption  of  coal  is  4200  Ibs.  a  day.     Find 
the  boiler  horse-power.  Ans.  54.739. 


CHAPTER  V. 
THE  VALVE  AND  ITS  MOTION. 

70.  In  the  steam  engine,  steam  is  admitted  to,  and  exhausted 
from,,  each  end  of  the  cylinder  by  a  valve  actuated  by  a  part  of  the 
engine  itself. 

With  the  exception  of  the  "  drop  "  valves  used  with  beam  engines 
of  slow  rotational  speed,  and  the  special  cases  of  rotating  valves, 
such  as  those  used  with  the  Corliss  engine,  the  ordinary  locomotive 
slide  valve,  and  its  modification  known  as  the  piston  valve,  are 
exclusively  used  in  stationary  and  marine  practice.  The  inadapta- 
bility of  the  drop  and  rotating  valves  to  engines  of  high  rotational 
speeds,  due  to  their  trip  gear,  has  limited  their  use. 


The  smooth  and  economical  working  of  an  engine  depends,  to  a 
large  extent,  upon  the  proper  admission  and  release  of  steam  to  and 
from  the  cylinder,  and  it  is,  therefore,  very  important  that  study 
should  be  given  to  valves  and  their  design. 

In  Figs.  2,  3,  4,  5,  6,  and  7,  the  hatched  sections  represent  the 
valve  seat  and  parts  of  the  cylinder  anrl  piston,  while  the  solid  sec- 
tion is  that  of  an  ordinary  single-ported  slide  valve. 

In  the  valve  seat  there  are  three  passages,  or  ports,  two  leading 
from  the  steam,  or  valve,  chest  to  the  cylinder,  and  the  other  lead- 
ing either  to  the  condenser,  the  atmosphere,  or,  in  the  case  of 
multiple-expansion  engines,  to  the  next  cylinder  in  the  order  of  the 
expansion.  The  steam  ports  leading  to  each  end  of  the  cylinder  are 
marked  8  and  8'9  and  the  exhaust  port,  through  which  the  steam 
leaves,  is  marked  E.  The  flat  face  of  the  valve  works  steam-tight 
on  the  flat  seat  on  the  side  of  the  cylinder. 


46 


THE  ELEMENTS  OF  STEAM  ENGINEERING 


The  motion  of  the  slide  valve  in  distributing  steam  in  the  cylin- 
der during  the  stroke  of  the  piston  P  will  be  understood  by  reference 
to  the  figures,,  the  arrows  indicating  the  direction  of  motion  of  the 
valve  and  piston. 

In  Fig.  2,  the  piston  is  just  commencing  its  stroke  to  the  right, 
the  valve  showing  a  slight  opening  of  the  port  at  8  for  the  admis- 
sion of  steam  into  the  cylinder.  This  slight  opening  of  the  port 


for  steam  when  the  piston  commences  its  stroke,  is  known  as  the 
"  lead "  of  the  valve,  and  is  always  provided  for.  The  entering 
steam  starts  the  piston  on  its  stroke,  the  valve  moving  in  the  same 
direction.  The  port  S'  has  been  opened  by  the  inside  edge  of  the 
valve,  thus  allowing  the  steam  in  the  cylinder,  which  has  been  used 
on  the  preceding  stroke  of  the  piston  to  the  left,  to  escape  through 


the  exhaust  port  E  and  exhaust  pipe  shown  in  circle.  The  valve 
and  piston  continue  their  motions  in  the  same  direction  until  the 
valve  reaches  the  limit  of  its  travel  to  the  right,  as  shown  in  Fig.  3, 
and  comes  for  an  instant  to  rest,  the  piston,  however,  continuing  its 
motion.  The  valve  being  now  at  its  extreme  point  of  travel  to  the 
right,  it  will  be  seen  that  the  port  S'  is  wide  open  to  the  exhaust, 
but  that  the  port  8  is  not  wide  open  for  the  admission  of  steam. 


THE  VALVE  AND  ITS  MOTION  47 

In  the  design  of  the  valve  it  is  always  arranged  to  have  a  full  open- 
ing of  the  port  for  exhaust,  but  a  full  opening  of  the  port  for  steam 
is  rarely  provided  for. 

The  eccentric,  which  is  the  actuating  mechanism,  now  causes  the 
valve  to  commence  its  return  stroke,  the  valve  and  piston  moving  in 
opposite  directions.  A  very  important  position  of  the  valve  is 
shown  in  Fig.  4,  when  the  steam  edge  of  the  valve  has  just  closed 
the  port  8  and  any  further  admission  of  steam  to  the  cylinder  is 
"  cut-off,"  the  remainder  of  the  stroke  of  the  piston  being  accom- 
plished by  the  expansive  force  of  the  steam  already  admitted.  The 
port  8',  it  will  be  noticed,  is  still  open  to  exhaust.  The  valve  and 
piston  continue  their  motions  in  opposite  directions,  and  Fig.  5 
shows  the  valve  in  mid-position,  the  piston  having  nearly  completed 
its  stroke,  and  the  exhaust,  or  inside,  edges  of  the  valve  coinciding 
with  the  inner  edges  of  the  steam  ports  8  and  8'.  This  latter  con- 


dition results  from  there'  being  no  inside,  or  exhaust,  lap  to  the 
valve,  which,  it  will  be  seen  later,  is  rarely  the  case.  With  the  valve 
in  this  position,  two  events  of  importance  occur.  The  closure  of 
the  port  S'  has  entrapped  within  the  cylinder  some  of  the  exhaust 
steam  of  the  preceding  stroke,  and  as  the  piston  proceeds  this  steam 
is  compressed,  its  pressure  gradually  increasing  as  the  piston  nears 
the  end  of  its  stroke.  This  "  compression "  provides  an  elastic 
cushion  of  steam  which  absorbs  the  momentum  of  the  reciprocating 
parts  of  the  engine,  and  brings  them  to  rest  without  shock.  The 
compressed  steam  assists  the  entering  steam  in  starting  the  piston 
on  the  return  stroke ;  and,  as  it  fills  the  clearance  space,  it  has  an 
appreciable  effect  on  the  amount  expended,  as  a  less  quantity  need 
be  drawn  from  the  boiler  at  each  stroke. 

The  left  hand  inner  edge  of  the  valve  having  arrived  at  the  inner 
edge  of  the  port  8,  the  other  event  now  occurs.     The  continued  mo- 


48 


THE  ELEMENTS  OF  STEAM  ENGINEERING 


tion  of  the  valve  in  the  direction  of  the  arrow  opens  the  port  S, 
and  the  steam  that  has  been  driving  the  piston  is  allowed  to  escape 
through  the  exhaust  port  E,  the  pressure  on  the  left  of  the  piston 
being  suddenly  reduced.  This  event  is  known  as  the  "  release  "  of 
the  steam. 

The  motions  in  opposite  directions  of  the  valve  and  piston  con- 
tinue, the  port  S  opening  wider  to  exhaust  and  the  pressure  of  the 


compressed  steam  increasing,  until  just  before  the  arrival  of  the 
piston  at  the  end  of  the  stroke,  when  the  outside,  or  steam,  edge  of 
the  valve  reaches  the  outside  edge  of  the  port  Sf,  as  shown  in  Fig.  6. 
Now  follows  the  event  of  "  admission,"  and  as  the  port  8'  uncovers, 
fresh  steam  is  admitted,  raising  the  compressed  steam  to  full  pres- 
sure, which  acts  in  opposition  to  the  piston  during  the  remainder  of 
the  stroke,  and  brings  it  gradually  to  rest  without  shock  at  the  end 
of  the  stroke. 


Fig.  7  shows  the  position  of  the  valve  and  piston  when  the  stroke 
is  completed,  the  port  S'  being  open  to  the  extent  of  the  "  lead." 
All  the  events  just  described  are  repeated  when  the  piston  makes 
the  return  stroke,  at  the  completion  of  which  the  valve  will  be  found 
in  the  position  shown  in  Fig.  2. 

In  this  manner  the  piston  is  given  a  reciprocating  motion  in  the 
cylinder,  which  is  communicated  to  the  shaft  by  jneans  of  the  con- 


THE  VALVE  AND  ITS  MOTION 


49 


necting-rod  and  crank  mechanism,  and  the  shaft  revolves  as  long  as 
steam  is  supplied  to  the  cylinder. 

71.  The  Eccentric. — Slide  valves  are  usually  actuated  by  an  ec- 
centric and  rod.  The  eccentric  consists  of  a  disc  keyed  to  the  shaft 
and  revolving  with  it,  but  having  its  center  eccentric  to  that  of  the 
shaft.  In  Fig.  8,  C  is  the  center  of  the  shaft,  and  E  the  center  of 
the  eccentric.  The  eccentric  disc  is  usually  in  two  parts,  D  and  D' ', 
bolted  together  as  indicated.  As  the  shaft  revolves,  the  eccentric 
is  carried  with  it,  the  latter  turning  within  the  strap,  thus  giving 
to  the  end  A  of  the  rod  a  reciprocating  motion  which  is  communi- 
cated to  the  valve.  The  end  A,  which  is  guided  to  move  in  a  straight 
line  pointing  to  the  center  of  the  shaft,  is  directly  connected  to 
the  valve  stem  in  engines  designed  to  run  only  in  one  direction, 
but  in  locomotives  and  marine  engines,  where  the  direction  of  mo- 


8. 


tion  is  often  reversed,  the  reciprocating  motion  is  communicated 
through  the  agency  of  a  link. 

An  examination  of  Fig.  8  will  show  that  the  motion  of  the  end 
A  of  the  rod  along  the  line  CB  will  be  twice  the  distance  CE.  CE 
is  the  distance  between  the  centers  of  the  shaft  and  eccentric.  It 
is  variously  called  the  "  throw  of  the  eccentric,"  "  eccentric  arm," 
and  "eccentric  radius,"  and  it  is  equal  to  the  half-travel  of  the 
valve. 

It  will  readily  be  seen  that  the  motion  given  to  the  point  A  is 
exactly  equivalent  to  that  which  would  be  produced  by  a  crank  CE 
and  connecting  rod  EA,  from  which  we  conclude  that  the  eccentric 
and  rod  is  a  special  case  of  the  connecting-rod  and  crank  mechanism, 
in  which  the  crank  pin  is  made  so  large  as  to  include  the  shaft 
within  its  section.  In  order  that  this  may  be,  the  radius  of  the 
4 


50  THE  ELEMENTS  OF  STEAM  ENGINEERING 

eccentric  must  be  greater  than  the  sum  of  the  radii  of  the  crank  and 
of  the  shaft;  that  is,  Erf  >  EC  +  Cr  (Fig.  8). 

72.  Having  considered  the  action  and  actuating  mechanism  of 
the  slide  valve  in  the  distribution  of  the  steam  in  the  cylinder,  it  is 
next  in  order  to  fix  the  position  of  the  eccentric  on  the  shaft,  rela- 
tive to  the  crank,  so  that  the  events  illustrated  by  Figs.  2,  3,  4,  5, 
6,  and  7  shall  occur  at  the  proper  times. 

For  this  purpose  a  slide  valve  with  faces  of  just  sufficient  width 
to  cover,  when  in  central  position,  the  steam  ports  S  and  S',  Fig.  9,. 
will  be  used  in  illustration. 

In  order  that  alternate  strokes  of  the  piston  in  the  cylinder  shall 
be  in  opposite  directions,  the  valve  must  provide  that,  at  each 
stroke,  steam  shall  be  admitted  to  the  driving  side  of  the  piston , 
and  the  steam  used  in  the  preceding  stroke  shall  be  exhausted  from 


*• 


the  opposite  side.  The  valve  in  Fig.  9  is  in  its  central  position, 
and  if  its  movement  were  so  timed  that  it  would  occupy  that  position 
simultaneously  with  the  arrival  of  the  piston  at  the  end  of  either 
stroke,  the  provision  required  of  the  valve  would  be  fulfilled ;  pro- 
vided the  movement,  or  travel,  of  the  valve,  during  one  stroke  of  the 
piston,  were  equal  to  twice  the  width  of  the  steam  port,  and  that 
the  first  half  of  the  valve  movement  were  in  the  same  direction  as 
that  of  the  piston,  and  the  last  half  in  the  contrary  direction.  Since 
the  throw  of  the  eccentric  is  equal  to  the  half-travel  of  the  valve, 
and  as  the  crank  moves  with  the  piston,  it  is  at  once  evident  that 
the  desired  motion  to  the  valve  of  Fig.  9  would  be  obtained  by  plac- 
ing the  eccentric  on  the  shaft  90°  ahead  of  the  crank. 

73.  We  will  now  consider  the  distribution  of  the  steam  in  the 
cylinder  by  the  valve  of  Fig.  9,  the  eccentric  being  placed  on  the 
shaft  90°  ahead  of  the  crank.  The  piston  being  at  the  end  of  the 
stroke,  the  eccentric  will  cause  the  valve  to  move  to  the  right  to 


THE  VALVE  AND  ITS  MOTION 


51 


admit  steam  to  the  cylinder  through  the  port  8  to  drive  the  piston 
in  the  same  direction.  Simultaneously  with  the  admission  of  steam 
into  8,  the  right-hand  inside  edge  of  the  valve  will  open  the  port  8' 
to  the  exhaust,  and  there  will  not  have  been  any  "  release  "  of  the 
steam  before  the  piston  arrived  at  the  end  of  the  preceding  stroke, 
nor  was  there  any  "  lead  "  to  the  valve  to  give  the  full  steam  pres- 
sure on  the  piston  at  the  moment  of  commencing  the  stroke  under 
consideration.  After  the  eccentric  arm  has  revolved  through  90°, 
the  valve  will  have  reached  its  extreme  distance  to  the  right,  a  dis- 
tance just  sufficient  to  open  wide  the  port  8  to  steam  and  the 
port  S'  to  exhaust.  The  crank  and  eccentric  continue  to  revolve, 
the  piston  proceeds  on  its  stroke  •  but  the  instant  the  eccentric 
passes  the  90°  point  of  its  rotation,  the  valve  commences  its  travel 


to  the  left,  and  after  a  further  angular  motion  of  90°  by  the  crank 
and  eccentric,  the  piston  will  have  arrived  at  the  end  of  its  stroke 
simultaneously  with  the  arrival  of  the  valve  to  its  original  position. 
Thus,  it  is  seen,  there  has  been  no  "  cut-off  "  of  the  steam  before 
the  arrival  of  the  piston  at  the  end  of  its  stroke,  and,  therefore,  no 
advantage  taken  of  the  steam's  expansive  force  to  complete  the 
stroke;  nor  has  the  exhaust  closed  before  the  completion  of  the 
stroke  to  provide  for  that  very  necessary  elastic  cushion  of  com- 
pressed steam  to  gradually  bring  the  piston  to  rest  without  shock. 

74.  The  relative  movements  of  the  crank  and  eccentric,  as  just 
explained,  will  be  better  understood  by  referring  to  Fig.  10.  CR 
and  CE  are  the  original  positions  of  the  crank  and  eccentric  arms, 
the  position  of  the  valve  being  as  shown  in  Fig.  9.  CR'  and  CE'  are 


52  THE  ELEMENTS  OF  STEAM  ENGINEERING 

the  crank  and  eccentric  positions  after  the  shaft  has  revolved 
through  90°  the  valve  having  moved  its  extreme  distance  CE'  to 
the  right,  opening  the  ports  wide  to  steam  and  exhaust.  A  further 
rotation  of  90°  by  the  shaft  brings  the  crank  and  eccentric  to  the 
positions  CE"  and  CE",  respectively,  the  valve  having  returned 
through  a  distance  E'C  to  its  original  position,  just  as  the  piston 
completed  its  stroke;  for  while  the  center  of  the  crank  pin  has 
rotated  through  the  semicircle  RR'R",  its  motion  of  translation  has 
been  the  diameter  RE"  of  the  crank  pin  circle,  which,  of  course,  is 
equal  to  the  stroke  of  the  piston.  If  the  shaft  were  to  complete  its 
revolution,  as  indicated  by  the  dotted  semicircles,  the  crank  and 
eccentric  would  assume  their  original  positions  CR  and  CE,  and 
the  valve  would  have  moved  its  extreme  distance  CE'"  to  the  left 
and  returned  the  same  distance  E'"C  to  its  original  position,  the 
piston,  meanwhile,  completing  its  return  stroke,  and  the  positions 
of  the  valve  and  piston  would  be  as  represented  in  Fig.  9. 

The  action  of  this  valve  is  clearly  defective,  and  the  difficulty 
results  from  delaying  until  the  end  of  the  stroke  the  changes  which 
are  necessary  for  its  reversal. 

In  practice  this  difficulty  is  avoided  by  making  the  valve  longer, 
so  that  when  in  its  central  position,  its  faces  will  more  than  cover 
the  steam  ports,  as  shown  in  Fig.  5 ;  and  in  addition  to  this  length- 
ening of  the  valve,  the  eccentric  is  set  on  the  shaft  an  angular  dis- 
tance greater  than  90°  ahead  of  the  crank. 

Definition. — Lap  of  the  valve — called  outside,  or  steam,  lap — is 
the  distance  the  outer,  or  steam,  edge  of  the  valve  extends  beyond, 
or  laps  over,  the  steam  edge  of  the  port,  when  the  valve  is  in  the 
mid-position  of  its  stroke. 

It  has  been  pointed  out  that  giving  lead  to  the  valve  provides 
that  a  full  pressure  of  steam  shall  be  exerted  on  the  piston  at  the 
commencement  of  its  stroke,  and  that  assistance  be  rendered  the 
compressed  steam  in  bringing  the  piston  momentarily  to  rest  with- 
out shock  at  the  end  of  the  stroke. 

Definition. — Lead  of  the  valve  is  the  distance  the  valve  uncovers 
the  port  for  the  admission  of  steam  when  the  piston  is  at  the  end 
of  its  stroke. 

Eegarding  the  valve  of  Fig.  5  as  that  of  Fig.  9  lengthened  to 
overlap  the  ports,  it  will  be  seen  that  in  order  to  give  it  lead  when 


THE  VALVE  AND  ITS  MOTION  53 

the  piston  is  at  the  end  of  its  stroke,  as  shown  in  Fig.  2,  page  45, 
we  must  first  advance  the  eccentric  CE,  Fig.  11,  through  an  angle 
EGA,  called  the  angle  of  lap,  to  move  the  valve  a  distance  CP  equal 
to  the  lap,  and  then  advance  it  through  the  angle  ACB,  called  the 
angle  of  lead,  to  move  the  valve  a  distance  PD,  equal  to  the  de- 
sired lead. 

Definition. — Angular  Advance  of  the  eccentric  is  the  number  of 
degrees  in  excess  of  a  right  angle  that  the  eccentric  is  set  on  the 
shaft  in  advance  of  the  crank,  and  is  equal  to  The  Lap  Angle  -\-  The 
Lead  Angle. 

75.  The  primary  object  in  giving  the  lap  to  a  valve  is  to  provide 
means  for  cutting  off  the  admission  of  steam  into  the  cylinder  be- 
fore the  end  of  the  stroke,  so  that  advantage  may  be  taken  of  the 
expansive  power  of  the  steam  to  continue  the  movement  of  the 
piston  to  the  end  of  the  stroke.  There  is  a  limit,,  however,  to  the 


Fiq.  II 

amount  of  lap  that  should  be  given  a  valve.  The  addition  of  lap 
necessitates  an  increase  of  the  angular  advance  of  the  eccentric 
in  order  to  maintain  the  lead,  and  if  the  maximum  port  opening  is 
also  to  be  maintained,  the  travel  of  the  valve  must  be  increased.  It 
is  always  desirable  to  have  a  small  valve  travel  so  that  the  work 
expended  in  moving  the  valve  shall  be  a  minimum,  and  it  is  for 
this  reason  that  double-ported  valves  are  resorted  to,  as  by  their  use 
only  one-half  the  travel  of  the  single-ported  valve  is  necessary  for 
a  given  port  opening.  The  increase  in  the  angular  advance  of  the 
eccentric  will  have  the  effect  of  hastening  all  the  operations  of  the 
valve,  and  to  whatever  extent  the  opening  of  the  exhaust  has  been 
hastened  its  closure  will  likewise  be  hastened,  and  excessive  com- 
pression may  result.  With  a  fixed  eccentric  the  limit  of  cut-off  by 
means  of  lap  is  about  f  of  the  stroke. 

76.  If,  when  the  valve  is  in  its  mid-position,  the  inside  edges 
extend  over  the  inside  edges  of  the  ports,  the  valve  is  said  to  have 
inside,  or  exhaust,  lap,  and  its  effect  is  to  delay  the  release  of  the 


54  THE  ELEMENTS  OP  STEAM  ENGINEERING 

steam  and  hasten  the  compression.  When  in  its  mid-position,  if 
the  inside  edges  of  the  valve  do  not  overlap  the  inside  edges  of  the 
ports,  but  show,  instead,  an  opening  to  the  exhaust,  the  valve  is 
then  said  to  have  negative  exhaust  lap,  and  this  is  not  infrequently 
the  case,  particularly  with  vertical  engines  where  a  prompt  exhaust 
of  the  steam  from  the  cylinder  on  the  up  stroke  is  desirable.  An 
examination  of  Fig.  5  will  show  that  with  negative  exhaust  lap, 
there  will  be  a  period  when  both  ends  of  the  cylinder  will  be  in 
communication  with  the  exhaust  and  with  each  other.  This  will 
occur  just  before  compression,  and  the  released  steam  from  the 
driving  side  of  the  piston  will  flow  into  the  exhausted  side,  causing 
an  increase  in  the  back  pressure,  which  will  be  shown  by  a  rise  in 
the  back  pressure  line  of  the  indicator  diagram  just  before  compres- 
sion. The  period  of  this  communication  will  be  brief,  as  the  mo- 
tion of  the  valve  is  then  at  its  quickest. 

77.  The  necessity  for  giving  lead  to  the  valve  has  been  shown, 
and  its  amount  is  fixed  arbitrarily,  according  to  the  type  of  engine 
and   the   judgment   of    the    designer.     The    rotational    speed    in 
all  engines  arid  the  weight  of  the  reciprocating  parts  compared  to 
piston  area  in  vertical  engines,  are  governing  features  in  the  deter- 
mination of  the  amount  of  the  lead.     For  slow-running  stationary 
engines  the  lead  varies  from  -fa.  to  T*g-  of  an  inch.     For  the  high- 
speed stationary  engine  it  is  seldom  less  than  •£$  of  an  inch,  and  for 
the  locomotive  it  is  commonly  J  inch  when  at  full  speed  and  linked 
up.     The  weight  of  the  reciprocating  parts  of  a  vertical  engine  act 
with  the  steam  on  the  top,  or  down,  stroke,  and  against  the  steam 
on  the  bottom,  or  up,  stroke,  and  therefore  the  necessity  for  in- 
creased lead  at  the  bottom,  or  crank,  end  of  the  valve,  to  assist  com- 
pression in  bringing  the  piston  to  rest  without  shock  and  to  furnish 
promptly  a  full  pressure  to  start  the  piston  on  the  up  stroke.     As 
much  as  f  of  an  inch  for  the  bottom  lead  of  a  vertical  marine 
engine  is  not  uncommon. 

78.  For  reasons  which  will  be  explained  later,  the  angularity  of 
the  connecting-rod  introduces  an  inequality  in  the  movement  of  the 
piston,  which  occasions  a  greater  piston  displacement  on  the  for- 
ward (toward  the  shaft)  stroke  than  on  the  return  stroke,  for  equal 
angular  positions  of  the  crank,  and  this  inequality  is  greater,  the 
greater  the  ratio  of  the  length  of  the  crank  to  the  length  of  the 
connecting-rod.     Then,  since  the  cut-off  of  both  strokes  is  effected 


THE  VALVE  AND  ITS  MOTION  55 

by  the  same  eccentric,  there  will  be  an  inequality  in  the  two  cut-offs, 
later  on  the  forward  than  on  the  return  stroke.  In  stationary  en- 
gines, where  the  crank-connecting-rod  ratio  is  comparatively  small, 
this  inequality  is  not  of  much  consequence,  and  a  partial  compen- 
sation is  usually  provided  by  giving  a  trifle  less  lap  on  the  crank 
end  of  the  valve  than  on  the  head  end,  without  seriously  disarrang- 
ing the  leads.  To  provide  for  an  absolute  equality  in  cut-off 
would  require  so  little  lap  on  the  crank  end  as  to  make  the  lead 
excessive.  In  vertical  marine  engines,  the  crank-connecting-rod 
ratio  is  comparatively  large,  and  the  inequality  in  cut-off  more  pro- 
nounced; but  since,  as  already  stated,  a  large  lead  is  desirable  on 
the  bottom,  or  crank,  end,  the  lap  on  that  end  may  be  sensibly  less 
than  on  the  top  end,  which  will  afford  a  partial  equalization  of  the 
cut-off.  In  addition  to  this,  it  is  common  marine  practice  to  make 
the  top  cut-off  a  trifle  later  than  that  desired,  which  will  make  the 
bottom  cut-off  also  later,  and  therefore  the  mean  cut-off  more  nearly 
that  desired. 

79.  Setting  Slide  Valves. — The  efficient  working  of  the  engine 
depends  largely  on  the  proper  adjustment  of  the  valve  on  its  stem, 
and  the  placing  of  the  eccentric  in  its  proper  place  on  the  shaft. 
The  performance  of  these  two  operations  is  called  "  setting  the 
valve."  It  is  first  necessary  to  locate  the  dead  points  of  the  engine, 
or  put  the  engine  "  on  its  center,"  as  it  is  called.  When  an  engine 
is  on  its  center,  a  line  joining  the  center  of  the  crank  pin  and  the 
center  of  the  shaft  is  in  direct  line  with  the  axis  of  the  cylinder, 
and  the  piston  is  then  absolutely  at  the  end  of  its  stroke.  To  ob- 
tain this  position  the  engine  is  jacked  until  the  cross-head  nearly 
reaches  the  end  of  its  stroke,  when  a  mark  is  made  on  the  slides  to 
correspond  to  one  made,  or  existing,  on  the  cross-head.  At  some 
point  on  the  engine  framing  near  the  crank  disc,  the  pulley,  or  the 
fly  wheel,  make  a  center  punch  mark,  and  with  this  mark  as  a  center, 
and  a  tram  of  convenient  length  as  a  radius,  describe  a  small  arc 
on  the  revolving  part  selected.  Now  jack  the  engine  past  the  end 
of  the  stroke  and  until  the  mark  on  the  cross-head  returns  to  the 
mark  made  on  the  slide.  From  the  center-punch  mark  as  a  center, 
describe  another  small  arc  with  the  tram  on  the  selected  revolving 
part.  The  cross-head  or  the  piston  is  now  the  same  distance  from 
the  end  of  the  stroke  as  it  was  when  the  first  arc  was  described  on 
the  revolving  part;  so  if  the  distance  between  these  two  arcs  is 


56  THE  ELEMENTS  OF  STEAM  ENGINEERING 

bisected,  and  the  point  of  bisection  marked  with  the  center  punch, 
then  when  the  engine  is  jacked  so  that  the  tram  exactly  spans  the 
distance  between  the  two  center-punch  marks,  the  engine  will  be 
on  the  center  and  the  piston  at  the  end  of  the  stroke.  A  like  pro- 
cess will  determine  the  other  dead  center.  In  putting  an  engine  on 
center,  and  during  the  whole  operation  of  setting  a  valve,  the  engine 
should  be  moved  in  the  direction  in  which  it  is  intended  to  run,  in 
order  that  all  lost  motion  may  be  taken  up  in  that  direction. 

Valves  are  usually  set  for  equal  leads,  the  exceptions  being  for 
vertical  engines,  as  already  pointed  out,  and  for  cases  where 
attempts  are  made  to  equalize  the  cut-off. 

To  set  a  slide  valve,  the  engine  is  placed  on  the  center,  and  if  the 
eccentric  be  not  fixed  on  the  shaft  in  its  proper  position  it  should 
be  approximately  so  placed,  making  the  angular  advance,  prefer- 
ably, a  trifle  larger  than  that  required.  By  means  of  the  nuts  on 
the  valve  stem  the  valve  is  given  the  proper  lead.  Turn  the  engine 
to  the  other  center,  and  if  the  lead  shown  there  is  not  the  same,  the 
difference  must  be  corrected — half  by  altering  the  length  of  the 
valve  stem,  and  half  by  moving  the  eccentric.  With  the  lead  thus 
equalized,  the  eccentric  should  be  keyed  fast  on  the  shaft  and  the 
nuts  on  the  valve  stem  tightened;  the  operation  will  then  be  com- 
plete. 

Should  the  valve  be  designed  for  unequal  leads,  the  method  of 
setting  it  is  exactly  the  same  as  just  described,  the  correction  being 
so  made  as  to  have  the  leads  as  desired. 

80.  The  Link  Motion. — With  marine  engines  and  locomotives, 
whose  direction  of  running  must  often  be  reversed,  the  reciprocat- 
ing motion  furnished  by  the  eccentric  is  communicated  to  the  valve 
through  the  agency  of  a  link.  The  link  motion  was  originally  de- 
signed as  a  means  of  reversing  the  motion  of  an  engine,  but  it  was 
afterwards  found  to  furnish  a  method  of  materially  increasing  the 
range  of  expansion  of  the  steam  in  the  cylinder. 

The  action  of  a  link  motion  in  reversing  an  engine  may  be  under- 
stood by  referring  to  Fig.  12,  where  CR  represents  the  crank  in 
some  one  position.  We  have  already  seen  that  in  order  to  have  the 
crank  revolve  in  the  direction  indicated  by  the  full  line  arrow,  the 
eccentric  must  be  fixed  on  the  shaft  at  some  position  CE  ahead  of 
the  crank.  Now  in  order  to  have  the  crank  turn  in  the  direction 
of  the  dotted  arrow,  the  eccentric  would  have  to  be  shifted  on  the 


THE  VALVE  AND  ITS  MOTION 


57 


shaft  to  the  position  CEf  ahead  of  the  crank  in  the  opposite  direc- 
tion. As  this  is  impracticable,  the  difficulty  is  overcome  by  having 
two  eccentrics  keyed  to  the  shaft,  one  having  its  center  at  E  and 
the  other  at  E'.  Each  eccentric  has  its  rod  connected  to  one  end 
of  a  curved  link  which,  as  originally  designed,  is  slotted  to  receive  a 


block  directly  connected  to  the  valve  stem,  and  by  a  suitable  arrange- 
ment of  levers  the  link  may  be  shifted  so  that  the  movement  of  the' 
valve  may  be  under  the  influence  of  either  the  go-ahead  or  the  back- 
ing eccentric  as  desired. 

81.  Figs.  13  and  14  are  skeleton  sketches  of  a  link  motion,  in 
which  CR  is  the  crank,  E  and  E'  the  centers  of  the  go-ahead  and 
backing  eccentrics,  respectively,  and  PQ  the  link,  curved  with  a 
radius  equal  to  the  length  of  the  eccentric  rod  EP. 

When  the  crank  is  pointing  away  from  the  link  and  the  eccentric 
rods  are  joined  to  the  ends  of  the  link  nearest  to  them,  as  in  Fig.  13, 
the  rods  are  said  to  be  "  open,"  and  if,  when  the  crank  is  in  the- 
same  position,  the  rods  are  joined  to  the  link  as  shown  in  Fig.  14,. 
the  rods  are  said  to  be  "  crossed."  If,  as  is  frequently  the  case  with 
piston  valves,  steam  is  taken  at  the  inside  edges,  the  rods  as  con- 
nected in  Fig.  13  would  be  called  "  crossed,"  and  if  as  in  Fig.  14r 
would  be  called  "  open." 

When  the  link  is  shifted  so  that  the  valve  block  5,  Fig.  13,  is 
brought  in  line  with  the  rod  EP,  it  is  in  full  forward  gear,  and 
the  motion  of  the  valve  is  governed  by  the  eccentric  E.  If  the  link 
be  shifted  so  that  the  block  comes  in  line  with  the  rod  E'Q,  it  is  in 


58 


THE  ELEMENTS  OF  STEAM  ENGINEERING 


full  backward  gear,  and  the  eccentric  E'  governs  the  motion  of  the 
valve.  When  the  block  is  midway  between  the  two  extreme  posi- 
tions the  link  is  said  to  be  in  mid-gear,  and  the  valve  is  influenced 
equally  by  both  eccentrics,  with  the  result  that  it  does  not  receive 
sufficient  motion  to  open  the  ports  and  the  engine  remains  at  rest. 


VAL  V£  ST£  M. 


Should  the  link  be  shifted  so  that  the  valve  block  is  brought  to  a 
position  intermediate  between  mid-gear  and  full  forward  gear,  as 
at  6r,  the  movement  of  the  valve  is  influenced  by  both  eccentrics, 
but  to  such  a  large  extent  by  eccentric  E  that  the  engine  will  con- 
tinue to  run  ahead.  The  effect,  however,  of  the  slight  influence  of 
the  backing  eccentric  has  been  to  decrease  the  travel  of  the  valve,  so 
that  all  its  operations  will'  be  earlier  than  when  working  in  full 
gear  and  the  cut-off  will  be,  therefore,  shorter. 


The  motion  given  to  the  valve-block  when  its  position  is  inter- 
mediate between  mid  and  full  gear  is  that  due  to  a  "  virtual "  eccen- 
tric of  less  throw  than  that  of  the  real  eccentric.  This  question 
has  been  investigated  by  designers,  and  it  has  been  found  that  for 
positions  of  the  valve  block  intermediate  between  mid  and  full  gear, 
the  centers  of  the  virtual  eccentrics  lie  in  a  parabolic  curve.  The 
locus  of  these  .centers  will  be  represented  with  sufficient  accuracy 
by  an  arc  of  a  circle  passing  through  the  centers  E  and  E'  of  the 


THE  VALVE  AND  ITS  MOTION  59 

eccentrics,  and  of  a  radius  equal  to  the  product  of  the  length  of 
the  eccentric  rod  and  half  the  distance  between  the  centers  of 
the  eccentrics,  divided  by  the  distance  between  the  eccentric  rod 

J71  P  V"    Tf  77" 

pins  in  the  link  =  -  ~2PQ •     ^is  ra-dius  ig  found  to  be  equal 

to  OE,  Figs.  13  and  14,  and  the  arc  EFE'  has  been  drawn,  concave 
to  the  center  of  the  shaft  for  open  rods,  and  convex  for  crossed  rods. 

To  find  the  virtual  eccentric  governing  the  motion  of  the  valve 
when  the  block  is  at  G,  divide  the  arc  EFE'  at  F  in  the  same  ratio 
that  G  divides  PQ ;  then  CF  is  the  throw,  and  DCF  the  angular 
advance  of  the  eccentric  which  is  virtually  actuating  the  valve. 
The  practical  way  of  determining  F  is  to  produce  PE  and  QE' 
until  they  intersect  at  some  point  R  in  the  center  line;  then  EG 
divides  the  arc  EE'  at  F  in  the  same  ratio  that  G  divides  PQ. 

The  action  of  the  link  in  providing  a  variable  cut-off  can  now  be 
understood. 

An  investigation  of  the  effects  of  the  open  and  crossed  rod  con- 
nections shows  that  with  open  rods  the  lead  increases  from  full  to 
mid  gear,  and  decreases  with  crossed  rods;  and  that  the  longer  the 
rods  are  made  and  the  shorter  the  link,  the  less  change  the  lead  will 
undergo.  It  is  the  usual  practice  to  make  the  open  rod  connection 
for  stationary  engines. 

82.  The  curvature  of  the  Stephenson  link  occasions  the  variation 
in  the  lead  from  full  to  mid  gear,  and  this  variation  becomes  greater 
as  the  eccentric  rods  are  shortened.     Zeuner  has  shown  analytically 
in  his  treatise  on  Valve  Gears  that  the  length  of  the  eccentric  rod 
should  be  the  radius  of  the  arc  of  the  link  in  order  that  the  leads 
may  be  equal.     It  may  be  shown,  too,  that  the  motion  of  the  valve 
becomes  more  nearly  harmonic  as  the  eccentric  rods  and  link  are 
lengthened.     Practice,  however,  has  about  fixed  the  length  of  the 
rods  to  about  twelve  times,  and  the  length  of  the  link  to  about  four 
times  the  throw  of  the  eccentric. 

83.  The  Stephenson  link  with  open  rod  connection  is  well  adapted 
for   locomotive   practice.     The   adjustment   should   be    such   that 
throwing  the  link  into  full  gear,  when  starting  the  train,  the  cut-off 
should  be  as  long  as  from  J  to  J  of  the  stroke  in  order  that,  with  a 
partially  open  throttle,  a  uniform  and  moderate  pressure  be  exerted 
on  the  piston  during  a  greater  part  of  the  stroke  to  overcome  the 
train  friction  without  causing  the  driving  wheels  to  slip.     This,  of 


60  THE  ELEMENTS  OF  STEAM  ENGINEERING 

course,  raises  the  terminal  pressure  and  is  the  cause  of  the  noisy 
exhaust  of  a  locomotive  when  starting  a  train. 

The  cranks  of  the  locomotive  being  set  at  right  angles  the  lead 
at  the  start  may  be  very  small,  but  as  the  speed  of  the  train  increases 
the  throttle  is  gradually  opened  and  then  the  cut-off  is  shortened 
and  the  lead  and  compression  increased  by  the  process  of  raising  the 
link.  The  locomotive,  being  a  high-speed  engine,  requires  a  con- 
siderable amount  of  compression  to  overcome  the  inertia  of  the 
reciprocating  parts. 

84.  The  crossed  rod  connection  is  commonly  used  for  marine 
engines,  for  then,  with  a  decreasing  lead  from  full  to  mid  gear,  the 
engine  will  stop  when  the  link  is  put  at  mid  gear,  which  it  would 
not  necessarily  do  with  the  rods  open  and  the  engine  running  with 
light  load. 

85.  The  slotted  Stephenson  link  is  expensive  to  make  and  diffi- 
cult to  hang  so  that  the  centers  of  the  eccentric  rod  pins  and  the 
center  of  the  valve  block  shall  be  in  line  and  the  block  work  easy 
in  the  slot.     These  difficulties  have  led  to  the  adoption  for  marine 
purposes  of  the  double-bar  link,  consisting  of  two  solid  bars  of 
rectangular  section  and  of  proper  curvature.     The  bars  are  secured 
together  by  distance  pieces  at  the  ends.     The  valve  block  is  between 
the  bars  and  has  flanges  which  bear  upon  the  tops  and  bottoms  of 
the  bars.     The  eccentric  rods  make  a  forked  end  connection  to  pins 
projecting  from  the  bars  near  the  ends. 

86.  The  Shifting  Eccentric  and  Automatic  Cut-Off. — As  stated 
on  page  36,  quick  running  of  an  engine  is  one  of  the  methods  of  de- 
creasing the  losses  from  condensation  and  re-evaporation,  and  it  is 
apparent  that  the  higher  the  rotational  speed  consistent  with  safe 
piston  velocity  the  greater  the  efficiency  of  the  steam  action  in  the 
cylinder;  and  since  piston  velocity  is  a  factor  of  the  power,  any 
increase  in  that  direction  admits  of  a  decrease  in  the  size  of  an 
engine  for  a  given  power.     Notwithstanding  these  advantages,  the 
introduction  of  the  high-speed  engine  was  long  delayed.     The  belief 
was  prevalent  that,  as  compared  with  one  of  lower  speed,  it  was 
more  liable  to  frequent  and  serious  accidents;  and  a  practical  ob- 
jection to  its  introduction  arose  from  the  fact  that  the  limit  of 
rotational  speed  was  soon  reached  at  which  the  "  drop "  cut-off 
could  be  applied,  and  it  became  essentially  necessary  for  the  intro- 
duction of  the  high-speed  engine  that  some  contrivance  be  devised 


THE  VALVE  AND  ITS  MOTION  61 

by  which  the  steam  should  have  a  positive  and  variable  cut-off,  auto- 
matically controlled,  in  order  that  the  speed  should  be  constant 
under  the  conditions  of  varying  load  and  pressure. 

The  vast  improvement  in  the  character  of  the  materials  of  con- 
struction; the  more  perfect  understanding  of  the  stresses  produced 
in  overcoming  the  inertia  of  the  reciprocating  parts;  the  skill  of 
the  American  workman — all  have  made  possible  the  production  of 
an  engine  of  such  perfect  balance  and  adjustment,  that  there  is  no 
longer  a  question  of  safety,  even  when  running  at  the  high  rotational 
speed  that  admits  of  a  direct  connection  to  the  armature  of  a 
dynamo-electric  machine. 

An  inherent  defect  in  the  "  fly-ball "  governor  precluded  its 
use  in  the  solution  of  the  cut-off  problem,  so  effort  was  centered  in 
devising  a  mechanism  in  which  the  centrifugal  action  of  revolving 
weights,  opposed  by  the  varying  tension  of  a  spring,  should  control 
the  action  of  the  valve  to  meet  the  exigency  of  a  sudden  variation 
in  the  load  on  the  engine  or  in  the  pressure  of  the  steam.  The 
result  was  the  production  of  what  is  known  as  the  Shaft  Governor, 
now  almost  exclusively  used  with  engines  driving  dynamo-electric 
machines  in  producing  electric  light  and  power. 

The  control  of  the  valve  by  the  shaft  governor  is  usually  through 
the  medium  of  a  shifting  eccentric,  and  while  the  high-speed  sta- 
tionary engines  of  rival  makers  differ  in  detail,  the  underlying  prin- 
ciples in  each  are  the  same. 

The  governor  is  usually  mounted  within  a  pulley  keyed  on  the 
crank  shaft,  and  its  motion  is,  therefore,  identical  with  that  of  the 
shaft.  The  eccentric  has  two  lugs  diametrically  opposite  each 
other,  from  one  of  which  it  is  suspended  in  the  center  line  of  the 
crank,  and  from  the  other  the  connection  is  made  with  the  linkage 
of  the  governor.  The  eccentric  is  slotted  so  that  its  center  may 
move  across  the  shaft. 

The  linkage  of  the  governor  being  properly  proportioned,  and 
the  weights  adjusted  for  a  definite  speed,  the  action  of  the  governor 
is  as  follows :  Any  decrease  in  load,  or  increase  in  pressure,  which 
would  tend  to  increase  the  speed  of  the  engine  and  the  centrifugal 
force  of  the  weights,  is  instantly  met  by  the  action  of  the  springs, 
which  moves  the  center  of  the  eccentric  nearer  the  center  of  the 
shaft,  thus  decreasing  the  throw  and  causing  an  earlier  cut-off.  An 
increase  of  load,  or  decrease  in  pressure,  causes  an  exactly  reverse 
action ;  and  thus  the  steam  supply  is  adjusted  to  the  load,  and  the 


62 


THE  ELEMENTS  OF  STEAM  ENGINEERING 


engine  kept  at  speed.  When  the  engine  is  at  rest  the  weights  are 
nearest  the  shaft  and  the  eccentric  is  at  its  greatest  throw,  corres- 
ponding to  the  full  gear  of  the  link  motion  :  as  the  engine  comes  up 
to  speed  the  travel  is  reduced  accordingly;  and  if  a  decreasing  load 
or  an  increasing  pressure  causes  a  still  further  reduction  of  travel, 
the  eccentric  approaches  its  minimum  throw  corresponding  to  mid 
gear  of  the  link  motion.  As  the  eccentric  moves  from  maximum  to 
minimum  throw  the  effect  on  the  lead  may  be  investigated  in  a 
manner  similar  to  that  employed  for  the  link  motion  (Figs.  13  and 
14). 

87.  Piston  Valve.  —  In  order  that  the  shaft  governor  shall  be 
sensitive  there  must  be  the  minimum  of  work  thrown  on  it,  and  this 


can  only  be  effected  by  having  the  least  possible  resistance  offered 
to  the  motion  of  the  valve.  This  clearly  points  to  the  use  of  a  bal- 
anced valve,  and  the  one  largely  used  is  a  variety  of  the  type  known 
as  piston-valve. 

Fig.  15  is  a  sectional  view  of  the  piston  valve  of  the  intermediate 
cylinder  of  the  triple-expansion  engine  in  the  mechanical  laboratory 
of  the  Baltimore  Polytechnic  Institute.  The  valve  may  be  con- 
ceived to  be  formed  from  the  ordinary  slide  valve  by  curving  .the 
latter  into  cylindrical  form.  It  is  arranged  to  take  steam  at  its 
inside  edges,  though  it  could  equally  well  be  arranged  to  do  so  at 
the  outside  edges.  It  will  be  seen  that  the  valve  is  balanced. 

When  made  for  large  marine  engines  the  pistons  of  the  valve  are 
fitted  with  springs  to  keep  them  steam  tight,  and  the  ports  are  cast 
with  diagonal  bars  to  keep  the  rings  from  springing  into  the  ports, 


THE  VALVE  AND  ITS  MOTION  63 

and  in  order  also  to  afford  a  continuous  guide  to  the  piston,  so  that 
the  rings  may  pass  over  the  ports  without  catching  on  the  edges. 

The  chief  disadvantage  of  the  piston  valve  is  the  difficulty  in 
keeping  it  steam-tight. 

88.  Radial  Valve  Gears. — Efforts  have  been  made  from  time  to 
time  to  supersede  the  use  of  the  link  motion  and  eccentrics  in 
actuating  the  valves  of  steam  engines,  and  of  these  the  most  suc- 
cessful have  been  the  radial  valve  gears  of  Marshall  and  Joy. 

In  the  Marshall  gear  the  link  is  dispensed  with  and  but  one 
eccentric  is  used.  With  the  Joy  gear  the  link  and  both  eccentrics 
are  dispensed  with,  the  valve  motion  being  obtained  from  a  point 
in  the  connecting  rod. 

Eadial  valve  gears,  when  accurately  proportioned,  provide  for  any 
degree  of  expansion  with  a  uniform  lead,  but  their  parts  are  neces- 
sarily heavy  and  difficult  of  correct  adjustment,  and  the  most  recent 
marine  practice  indicates  a  return  to  the  old  but  reliable  link  motion 
and  eccentric. 

PROBLEMS. 

24.  Diameter  of  shaft,  4";   diameter  of  eccentric,  10.5";   outside 
diameter  of  strap,  12";    distance  from  outside  to  outside  of  lugs, 
15.5";   distance  from  center  line  to  center  line  of  strap  bolts,  13"; 
width  of  lugs  of  each  strap,  1.5";  depth  of  groove  in  strap  to  retain 
eccentric,  0.25";  depth  of  enlargement  of  strap  where  rod  is  bolted, 
6"  at  right  angles  to  rod  and  1.75"  in  line  of  rod;   depth  of  rod  at 
strap  end,  2.5",  and  at  link  end,  1.25";   diameter  of  boss  at  end  of 
rod,  2.5";   diameter  of  eye  in  boss,  1";   length  of  rod,  13  times  .the 
throw  of  the  eccentric ;  distance  between  stud  bolts  at  strap  end  of 
rod,  4.5";    travel  of  valve,  5";    angular  advance,   38°.     Make  a 
sketch  of  the  eccentric  and  rod,  assuming  the  crank  to  be  on  the 
head  dead  center.     Scale,  1.5. 

25.  Throw  of  eccentric,  2";  length  of  eccentric  rods,  24" ;  length 
of  chord  of  link,  15";  angular  advance  of  the  eccentric,  38°  ;  stroke 
of  engine,  14".    Make  a  skeleton  sketch  of  the  link  motion,  and  find 
the  throw  of  the  virtual  eccentric  when  the  link  is  half  way  between 
full  and  mid  gear,  the  connection  of  the  rods  being  "  open."  Scale,  3. 

Ans.     Virtual  eccentric,  If". 


CHAPTER  VI. 

THE   CONVERSION   OF  MOTION.— ACTION   OF   THE   CRANK 
AND  CONNECTING  ROD. 

89.  If  the  point  'P9  Fig.  16,  moves  in  the  plane  of  the  paper  about 
•C  as  a  center,  it  will  describe  the  circle  RPB,  and  during  the  period 
of  describing  the  arc  EP,  the  circular  motion  of  P  may  be  consid- 
ered as  compounded  of  two  entirely  independent  movements  at 
right  angles  to  each  other — that  of  R  to  Q  in  the  line  of  the  diam- 
eter E8,  and  the  other  of  Q  to  P. 

Draw  the  diameter  AB  perpendicular  to  RS,  and  let  P  move  with 
.a  uniform  velocity.  If  we  denote  CP  by  r,  and  the  angle  PCR  by 

T 


0,  we  may  find  the  relation  between  the  positions  of  P  and  Q  as 
follows  : 

RQ  =  RC  —  QC  =  r  —  r  cos  6  —  r  (1  —  cos  0)  . 

To  find  the  position  of  Q  when  P  has  described  f  of  the  quad- 
rant RA. 

Here  we  have  9  =  60°,  and  cos  0  =  $,  consequently 


That  is,  P  describes  f  of  the  distance  circumferentially  from 
R  to  A  while  Q  moves  \  the  distance  RC  ;  and  since  the  motion  of 
P  is  uniform,  it  follows  that  the  last  half  of  RC  will  be  passed  over 
by  Q  in  one-half  the  time  that  it  required  to  pass  over  the  first 
half.  It  is  thus  seen  that  while  the  motion  of  P  is  uniform,  that 


THE  CONVERSION  or  MOTION  65 

of  Q  is  variable;  and  if  the  relative  positions  of  P  and  Q  were 
found  for  values  of  0  from  0°  to  180°  it  would  be  found  that  when 
P  is  at  R,  the  point  Q  is  at  rest,  and  as  P  continued  its  uniform 
motion,  the  velocity  of  Q  would  gradually  increase  until  the  value 
of  0  reached  90°,  when  the  velocity  of  Q  would  be  a  maximum  and 
equal  to  that  of  P;  as  P  moves  from  A  to  8,  Q  would  move  ,from 
C  to  8  with  diminishing  velocity  until  its  arrival  at  8,  when  it 
would  again  be  at  rest.  During  the  movements  just  described,  Q 
is  said  to  have  a  Simple  Harmonic  Motion. 

When  a  point  P  moves  uniformly  in  a  circle,  the  perpendicular 
PQ  let  fall  at  any  instant  to  a  fixed  diameter  R8,  intersects  the 
diameter  at  a  point  Q,  whose  position  changes  by  a  Simple  Har- 
monic Motion. 

To  ascertain  the  relation  existing  between  the  velocities  of  Q  and 
P,  draw  the  tangent  PT,  and  from  T  let  fall  the  perpendicular  TD 
upon  RS.  Suppose  the  velocity  of  P  to  be  such  that  in  a  given 
time  it  moves  a  distance  PT.  In  the  same  time  Q  would  move  to 
D,  and  we  shall  have : 

Velocity  of  Q  _  QD  _  PE  _  PQ  _.    . 

velocity  ^P~TT-"PT~TG~ 

The  sine  of  6  varies  from  0  to  1,  and  we  can  therefore  find  at 
any  period  of  the  motion  how  much  the  velocity  of  Q  differs  from 
that  of  P. 

90.  It  has  been  shown  how  the  valve  of  a  steam  engine,  actuated 
by  an  eccentric,  or  its  equivalent,  admits  steam  alternately  to  the 
two  ends  of  the  cylinder  to  drive  the  piston  forward  and  backward 
along  a  straight  path  of  definite  length  called  the  stroke.  This  re- 
ciprocating motion  of  the  piston  is  converted  into  one  of  continuous 
rotation  of  the  shaft,  through  the  agency  of  the  mechanism  of  the 
connecting  rod  and  crank.  In  this  mechanism  the  outer  end  of 
the  piston  rod  is  connected  to  a  cross-head,  which  is  constrained  by 
guides  to  move  in  line  with  the  axis  of  the  cylinder.  To  a  pin  in 
the  cross-head  the  inner  end  of  the  connecting  rod  is  so  attached  as 
to  permit  an  oscillating  motion  to  that  end  of  the  rod.  In  a  similar 
manner  the  outer  end  of  the  connecting  rod  is  connected  to  the 
crank  pin,  and  if  the  piston  be  driven  forward  or  backward,  the 
crank  will  be  caused  to  turn  about  the  axis  of  the  shaft  by  the  push 
or  pull  transmitted  from  the  cross-head  through  the  connecting  rod 
to  the  crank  pin. 
5 


66 


THE  ELEMENTS  OF  STEAM  ENGINEERING 


In  Fig.  17,  let  r  denote  the  length  CP  of  the  crank,  and  c  the 
length  PQ  of  the  connecting-rod.  If  from  R  and  8  we  lay  off  Ra 
and  Sb  each  equal  to  c,  then  ab  will  be  the  stroke  of  the  cross-head, 
and  for  purposes  of  discussion  it  is  the  stroke  of  the  piston,  the 
latter  being  rigidly  attached  to  the  cross-head.  Let  CP  be  a  posi- 
tion of  the  crank,  making  an  angle  6  with  the  center  line  Rb  of  the 
engine.  Then  Q  will  be  the  corresponding  position  of  the  piston, 
and  is  found  by  striking  an  arc  from  P  as  a  center  and  with  a 
radius  c.  With  Q  as  a  center  and  the  radius  c,  describe  the  arc  Pi. 
Then  t8  is  evidently  equal  to  QB;  and,  since  the  diameter  R8  of 
the  crank-pin  circle  is  equal  to  the  stroke  of  the  piston,  R8  may 
equally  well  represent  the  piston  stroke.  Then  t  conveniently  rep- 
resents the  position  of  the  piston  corresponding  to  the  crank  posi- 
tion CP,  and  it  is  at  once  evident  that  the  movement  of  the  piston 


is  not  the  result  of  a  simple  harmonic  motion.  The  angularity  of 
the  connecting-rod  introduces  the  inequality  tv  into  the  movement 
of  the  piston,  and  this  inequality  gradually  increases  from  its  zero 
value  at  the  dead  point  8  until  the  crank  arrives  at  the  position  CT, 
90°  in  advance  of  C8,  when  its  value  wC  is  a  maximum.  As  the 
crank  continues  its  rotation  the  inequality  gradually  diminishes 
until,  at  the  dead  point  R,  it  is  again  zero. 

An  examination  of  Fig.  17  shows  that  when  the  crank  is  at  the 
mid-point  of  its  revolution  from  8  to  R,  the  piston  has  traveled  a 
distance  Cw  beyond  its  mid-stroke;  and  it  is  seen  that  during  the 
first  quarter  of  the  revolution  on  the  forward  stroke  the  piston 
travels  a  greater  distance  than  during  the  second  quarter. 

This  inequality  can  be  avoided  only  by  giving  to  the  piston  a 
simple  harmonic  motion;  but  as  that  could  be  obtained  only  by 
making  the  connecting  rod  infinitely  long,  the  inequality  must 
always  exist  with  the  crank-connecting-rod  mechanism. 


THE  CONVERSION  OF  MOTION  67 

The  shorter  the  connecting-rod  the  greater,  of  course,  will  be  the 
inequality  in  the  movement  of  the  piston,  and  for  this  reason,  if 
for  no  other,  the  crank-connecting-rod  ratio  is  made  as  small  as 
possible.  In  marine  practice,  where  space  limits  the  length  of  the 
connecting-rod,  this  ratio  is  from  1  to  4  to  -1  to  5,  while  for  station- 
ary engines  it  is  not  uncommonly  as  small  as  1  to  7. 

A  further  examination  of  Fig.  17  shows  that  during  the  return 
stroke  of  the  piston,  conditions  exactly  the  reverse  of  those  of  the 
forward  stroke  obtain,  as  shown  by  the  dotted  lines. 

It  will  be  observed  that  the  use  of  the  connecting-rod  of  finite 
length  causes  the  cut-off  to  be  considerably  later  on  the  forward 
stroke  than  on  the  return  stroke,  and  the  reference  to  this  question 
on  page  54  can  now  be  understood. 

The  inequality  in  piston  displacement  for  the  same  crank  posi- 
tions of  the  two  strokes  results  in  an  unequal  distribution  of  steam 
in  the  cylinder  and,  therefore,  an  irregularity  in  the  driving  power 
of  the  engine.  The  angularity  of  the  connecting-rod  is  a  disturb- 
ing element  in  the  consideration  of  every  important  dynamic  ques- 
tion of  the  steam  engine,  rendering  more  or  less  difficult  the  solution 
of  problems  which  would  be  otherwise  simple. 

91.  With  the  aid  of  Fig.  17  we  may  find  an  expression  for  the 
relative  positions  of  the  crank  and  piston  at  any  instant. 

Let  a  denote  the  angle  made  by  the  connecting-rod  with  the  center 
line  of  the  engine,  and  denote  by  x'  the  distance  the  piston  has 
traveled  on  the  forward  stroke  when  the  crank  has  the  position  CP. 

x'  =  Cb  —  (Gv  +  vQ)  —  r  -f-  c  —  (r  cos  0  +  c  cos  a) 

=  r(l  —  cos  0)  +  c  —  c  cos  a  (1) 

We  have  h  =  r  sin  0,  and  cos  a  = 


therefore,  c  cos  a  =  \/c2  —  1?  =.  ^/c2  —  r2  sin2  0  . 
Substituting  the  value  of  c  cos  a  in  (1),  we  get  : 


x'  =  r(l  —  cos  0)  +  c  —  v'6'2  —  1*  sin 


=  r(l-co80)  +e^i_|y/i_^in'a 


68  THE  ELEMENTS  OF  STEAM  ENGINEERING 

If  x"  denote  the  distance  of  the  piston  from  the  crank  end  of  the 
stroke  for  the  same  crank  position,  we  shall  have  : 

x"  =  ab—b  =  2r-x' 


-  c  [l  -  Jl  -  ^' 


=  2r-r  +  rcostf 

=  r(l  Hh  eosfl)  -  .l-yl 
In  general  terms  we  shall  then  have  : 

(A) 


where  the  top  sign  must  be  taken  if  x  is  measured  from  the  head 
end  of  the  stroke,  and  the  lower  sign  if  measured  from  the  crank 
end. 

The  position  of  the  piston  for  any  given  crank  position,  or  con- 
versely, the  crank  angle  for  a  given  position  of  the  piston,  can  be 
found  by  equation  (A). 

The  ratio  r/c  will  become  very  small  if  raised  to  a  power  above 

the  square,  so  if  in  equation  (A)  the  expression      1  —    —  s  — 

be  expanded  by  the  binomial  theorem,  and  the  terms  containing 
higher  powers  of  c  than  the  square  be  rejected,  no  appreciable  error 
will  result,  and  the  formula  will  be  more  convenient  for  use. 
Thus, 


••.      ,  =,(!=,=  COB  fl)±.         .   :          •  (B) 

EXAMPLE  I.—  Let  c  =  40",  r  —  10",  0  =  210°  :  find  the  position 
of  the  piston,  measured  from  either  end  of  the  stroke. 

Here  we  have,  ^  =  |  ,  cos  210°  =  —  cos  30°  =  —  O.SG6,  and 

sin  210°  =  —  sin  30°  =  —  £. 

Substituting  in  (B)  we  have: 

x'  =  10(1  +  0.866)  +  rV  =  18-66  +  0.3125  ==  18.9725  inches 
when  measured  from  the  head  end  of  the  stroke;  and 


THE  CONVERSION  OF  MOTION  69 

x"  =  10(1  —  0.866)  —  TV  =  1.34  —  0.3125  =  1.0275  inches 
when  measured  from  the  crank  end  of  the  stroke. 

Equation  (B)  is  of  use  in  investigating  the  effect  of  the  obliquity 
of  the  connecting  rod  on  the  motion  of  the  piston.  For  practical 
purposes  the  relative  positions  of  the  crank  and  piston  may  be  ob- 
tained graphically. 

It  was  shown  on  page  64  that  r(l  —  cos  0)  would  be  the  value  of 
x  if  the  movement  of  the  piston  were  the  result  of  a  simple  har- 
monic motion,  but  as  that  could  be  obtained  only  with  a  connecting 

7*2  sin2  0 
rod  of  infinite  length,  the  term ~ in  equation  (B)   is  the 

measure  of  the  error  in  piston  displacement  due  to  the  angularity 
of  the  connecting  rod  of  finite  length. 

PEOBLEM. 

26.  Length  of  connecting-rod,  66";  stroke,  33".  Find  position 
of  the  piston,  measured  from  both  ends  of  the  stroke,  when  the 
crank  has  passed  through  an  angle  of  135°  on  the  head  stroke. 

Ans.  x'  =  29.1984".     x"  —  3.8016". 


CHAPTEE  VII. 
STAGE   EXPANSION   ENGINES. 

92.  A  stage  expansion  engine  is  one  so  designed  that  the  steam, 
after  entrance  into  one  cylinder  and  there  partially  expanded  in  the 
performance  of  work,,  is  exhausted  into  a  second,  third,  and  not 
infrequently  into  a  fourth  larger  cylinder,  for  further  expansion 
and  work  before  being  finally  exhausted  into  a  condenser. 

The  term  stage  expansion  is  used  to  avoid  the  confusion  which 
sometimes  arises  from  the  use  of  compound.  All  stage  expansion 
engines  are  compound,  but  by  common  consent  the  term  compound 
is  restricted  to  engines  in  which  the  expansion  takes  place  in  only 
two  cylinders,  or  in  two  stages,  while  the  terms  triple  expansion  and 
quadruple  expansion  are  applied  to  engines  in  which  the  expansion 
takes  place  in  three  and  four  cylinders  respectively.  The  first  and 
last  cylinders  in  the  order  of  the  expansion  are  called,  respectively, 
the  high-pressure  and  the  low-pressure  cylinders;  any  cylinder  or 
cylinders  which  intervene  are  known  as  intermediate  cylinders. 

93.  There  has  been  much  discussion  concerning  the  relative  merits 
of  the  simple  and  stage  expansion  types  of  engines,  but  the  results 
of  comparative  tests  clearly  indicate  an  economical  advantage  of 
from  10  per  cent  to  20  per  cent  in  favor  of  the  compound  system. 

Perhaps  the  most  convincing  evidence  of  the  advantage  in  the 
use  of  the  stage  expansion  engine  is  that  it  has  almost  entirely  dis- 
placed the  simple  engine  for  marine  purposes,  and  has  come  largely 
into  use  in  mill  engines,  locomotives,  and  in  stationary  plants  where 
large  powers  are  developed. 

It  has  been  shown  that  a  given  weight  of  steam  in  working  from 
a  higher  to  a  lower  pressure  is  capable  of  doing  a  definite  amount  of 
work,  and  while  the  number  of  cylinders  through  which  the  expan- 
sion takes  place  can  make  no  increase  in  this  theoretical  limit  of 
work,  it  does  make  a  very  material  difference  in  the  manner  of  its 
performance. 

A  prerequisite  to  the  use  of  the  stage  expansion  system,  and  the 
one  to  which  its  economy  is  solely  due,  is  a  high  pressure  of  steam ; 


STAGE  EXPANSION  ENGINES  71 

and  since  the  terminal  pressure  desired  is  the  same,  whatever  the 
system,  the  higher  the  pressure  the  greater  the  rate  of  expansion. 

It  was  pointed  out  on  page  38  that  the  attainment  of  a  high  rate 
of  expansion  in  a  single  cylinder  would  be  accompanied  with  such 
a  wide  range  of  temperature  as  to  occasion  excessive  initial  conden- 
sation and  final  re-evaporation.  The  consequent  wide  variation  in 
pressure  would  lead  to  very  objectionable  irregularity  of  rotational 
effort  on  the  crank-pin  and  excessive  strains  on  the  framing. 

With  a  stage  expansion  engine  this  high  rate  of  expansion  would 
be  distributed  through  two  or  more  cylinders,  thus  greatly  reducing 
the  range  of  temperature  and  pressure  in  each,  resulting  in  a  reduc- 
tion of  condensation  and  the  production  of  a  more  uniform  turning 
moment. 

94.  There  are  two  classes  of  stage  expansion  engines,  known  as 
the  "  continuous  expansion  "  and  the  "  receiver  "  types.    The  essen- 
tial difference  is  that  with  the  continuous  expansion  type  the  cranks 
must  be  placed  with  each  other  or  directly  opposite  each  other  on 
the  shaft,  while  with  the  receiver  engine  the  cranks  may  be  placed 
at  any  angle  with  each  other. 

The  continuous  expansion  engine  is  one  of  the  two-stage  type  in 
which  the  steam  enters  the  low-pressure  cylinder  as  fast  as  it  is  ex- 
hausted from  the  high-pressure  cylinder,  a  familiar  example  of 
which  is  that  of  the  tandem  engine  where  the  two  pistons  are  on 
one  rod. 

The  receiver  engine  derives  its  name  from  the  fact  that  originally 
it  was  designed  to  have  an  intermediate  reservoir  to  receive  the 
steam  as  it  was  exhausted  from  the  high-pressure  cylinder,  and  in 
which  it  remained  until  the  valve  opened  to  admit  it  into  the  low- 
pressure  cylinder.  The  size  of  this  receiver  equalled  that  of  the 
high-pressure  cylinder,  and  even  larger.  This  separate  reservoir 
was  found  to  be  unnecessary,  inasmuch  as  the  low-pressure  steam 
chest,  the  exhaust  pipe  from  one  cylinder  to  the  other,  and  the  por- 
tion of  the  high-pressure  cylinder  yet  filled  with  steam  when  the 
low-pressure  valve  opened,  were  found  to  furnish  ample  receiver 
space. 

95.  The  common,  form  of  the  compound,  or  two-stage  expansion, 
engine  with  cranks  at  right  angles  is  shown  in  Fig.  18.     When  this 
form  of  engine  is  designed  for  such  large  powers  as  require  a  low- 
pressure  cylinder  greater  than  100  inches  in  diameter,  the  three- 


72  THE  ELEMENTS  OF  STEAM  ENGINEERING 

cylinder  compound  'engine  with  two  low-pressure  cylinders,  as  shown 
in  Fig.  19,  should  be  adopted.  In  this  case  the  excessively  large 
low-pressure  cylinder  is  replaced  by  two  cylinders  whose  combined 
volume  equals  that  of  the  large  cylinder  they  replace,  and  notwith- 


standing  the  increased  space  occupied,  and  the  greater  number  of 
parts  required  with  no  increase  in  the  power  developed,  this  arrange- 
ment is  very  advantageous,  as  the  pieces  are  lighter  and  more  easily 
made,  and  the  increase  of  the  number  of  cranks  from  two  to  three 


enables  them  to  be  placed  at  angles  of  120°  on  the  shaft,  thus  insur- 
ing reduced  straining  and  a  steadier  motion  on  the  engine.  With 
this  engine  the  work  may  be  equally  divided  between  the  three  cylin- 
ders— one-third  being  allotted  to  each — by  considerably  increasing 


STAGE  EXPANSION  ENGINES  73 

the  receiver  pressure,  which  has  the  effect  of  reducing  the  power 
developed  in  the  high-pressure  cylinder  and  increasing  that  devel- 
oped in  each  of  the  low-pressure  cylinders. 

The  two-stage  compound  engine  is  advantageously  employed 
where  the  pressure  varies  from  80  to  120  Ibs.  absolute.  With  pres- 
sures varying  from  120  Ibs.  to  190  Ibs.,  the  three-stage,  or  triple, 
expansion  type  should  be  adopted,  and  it  is  the  type  which  is  very 
largely  used  in  the  merchant  marine  and  in  vessels  of  war.  For 
pressures  varying  from  190  Ibs.  to  275  Ibs.,  the  four-stage,  or  quad- 
ruple, expansion  type  is  used.  This  type  has  not  as  yet  come  into 
general  use,  and  its  employment  is  confined  to  small  vessels  where 
lightness  of  the  machinery  admits  of  high  rotational  speeds. 


The  common  arrangement  of  the  three-cylinder  triple-expansion 
engine  is  shown  in  Fig.  20.  The  first  receiver,  R,  leads  from  the 
exhaust  of  the  high-pressure  cylinder  to  the  valve  chest  of  the  inter- 
mediate cylinder,  and  the  second  receiver,  R',  connects  the  inter- 
mediate exhaust  with  the  low-pressure  valve  chest.  The  cranks  are 
placed  at  angles  of  120°  with  each  other,  and  their  sequence  on  the 
shaft  is  a  matter  of  dispute,  but  the  order  of  high,  low,  intermediate, 
seems  to  obtain  the  greatest  favor  from  the  fact  that  such  arrange- 
ment secures  a  more  uniform  receiver  pressure. 

96.  The  most  serious  inherent  defect  of  the  stage  expansion  sys- 
tem is  the  loss  occasioned  by  "  drop."  "  Drop  "  means  simply  the 
fall  in  pressure  between  the  terminal  pressure  in  one  cylinder  and 
the  initial  pressure  in  the  next  succeeding  cylinder  in  the  expan- 


74  THE  ELEMENTS  OF  STEAM  ENGINEERING 

sion.  This  is  occasioned  by  friction  in  the  exhaust  passages  and 
pipes,  and  by  the  unrestricted  expansion  in  the  receiver.  While  it 
is  impossible  to  avoid  entirely  the  loss  from  "  drop/'  measures  should 
be  taken  to  prevent  its  being  excessive.  Earlier  cut-offs  and  in- 
creased compression  in  the  cylinders  succeeding  the  high-pressure 
cylinder  have  a  tendency  to  minimize  "  drop." 

97.  The  mean  pressure  obtained  in  the  stage  expansion  engine 
with  a  given  rate  of  expansion  is  less  than  would  be  obtained  were 
the  expansion  all  to  take  place  in  one  cylinder,  and  this  difference  is 
due  to  "  drop."  The  steam,  when  its  expansion  is  completed,  occu- 
pies the  low-pressure  cylinder  only  and  were  there  no  loss  from 
"  drop  "  the  size  of  this  cylinder  for  a  given  power  would  be  the  same 
as  that  of  a  single-cylinder  engine  working  with  the  same  pressure 
and  the  same  .ratio  of  expansion.  Owing,  however,  to  the  loss  from 
"  drop  "  the  size  of  the  low-pressure  cylinder  must  be  made  some- 
what larger  than  would  suffice  for  the  single  cylinder  of  a  simple 
engine  of  the  same  power.  From  this  it  is  seen  that  the  high  and 
intermediate  cylinders  add  nothing  to  the  power  of  the  arrangement, 
and  that  the  low-pressure  cylinder  is  the  measure  of  the  power. 


CHAPTEE  VIII. 
THE   INDICATOR   AND   ITS   DIAGRAM. 

98.  The  steam  engine  indicator  is  an  instrument  devised  to  show 
primarily  the  steam  pressure  within  the  cylinder  at  every  point  of 
the  stroke. 

It  consists  essentially  of  a  small  steam  cylinder  containing  a 
piston  whose  vertical  movement,  when  acted  on  by  the  pressure  of 
the  steam  beneath  it,  is  opposed  by  a  spiral  spring  of  known  tension. 

The  movement  of  the  piston  actuates  a  parallel  motion  consisting 
of  a  system  of  light  levers,  a,  point  in  which  is  constrained  to  move 
in  a  straight  line  parallel  to  the  motion  of  the  piston.  This  parallel 
point  of  the  system  carries  a  pencil  which  reproduces  the  vertical 
movement  of  the  indicator  piston,  magnified  from  4  to  6  times  by 
means  of  the  system  of  levers.  In  addition  there  is  a  cylinder,  or 
drum,  to  which  a  paper  is  attached,  and  which  receives  a  forward 
and  backward  motion  of  rotation  on  its  own  axis;  the  forward 
motion  by  means  of  a  string  attached  to  the  cross-head  or  other  part 
of  the  engine  having  a  motion  exactly  coincident  with  that  of  the 
engine  piston,  and  the  backward  motion  by  means  of  the  reaction 
of  a  flat  coiled  spring  which  is  attached  to  the  base  of  the  drum, 
and  which  is  drawn  in  tension  by  the  forward  motion  of  the  drum. 

It  can  readily  be  understood  that  when  the  instrument  is  in  opera- 
tion and  the  pencil  pressed  against  the  paper,  the  combination  of 
the  vertical  motion  of  the  pencil  and  the  horizontal  motion  of  the 
drum  will  cause  a  closed  area  to  be  described  that  will  represent  the 
effective  work  done  by  the  steam  on  the  engine  piston. 

When  communication  is  opened  between  the  head  end  of  the 
engine  cylinder  and  the  lower  end  of  the  indicator  cylinder,  the 
vertical  motion  of  the  indicator  piston  during  the  forward  rotation 
of  the  drum  is  that  due  to  the  pressure  of  the  steam  in  the  engine 
cylinder  during  the  forward  stroke  of  the  piston;  and  during  the 
backward  rotation  of  the  drum  it  is  that  due  to  the  pressure  on  the 
same  side  of  the  piston  during  the  return  stroke.  The  upper  part 
of  the  diagram  is,  therefore,  a  representation  of  the  varying  pres- 
sure on  the  engine  piston  during  its  forward  stroke,  and  the  lower 
part  a  representation  of  the  back  pressure  opposed  to  the  piston  on 


76  THE  ELEMENTS  OF  STEAM  ENGINEERING 

its  return  stroke.  The  length  of  the  ordinate  of  the  diagram  at  any 
point  is,  therefore,  when  measured  to  the  scale  of  the  indicator 
spring,  the  effective  pressure  on  the  piston  at  that  point. 

The  closed  area,  or  diagram,  thus  described  represents  the  vary- 
ing pressure  on  only  one  side  of  the  engine  piston  during  one  revo- 
lution. When  communication  is  opened  between  the  indicator  and 
the  other  end  of  the  cylinder,  a  similar  diagram  will  be  obtained 
which  will  represent  the  varying  steam  pressure  on  the  other  side 
of  the  engine  piston  during  a  revolution. 

99.  Before  giving  an  illustration  of  a  diagram,  several  features 
of  the  indicator  will  be  noticed  in  order  that  the  application  of  the 
instrument  to  the  steam  engine  may  be  understood.  For  a  more 
detailed  description  any  one  of  the  many  publications  on  the  sub- 
ject may  be  consulted. 

The  pistons  of  the  indicators  in  ordinary  use  are  exactly  one-half 
square  inch  in  area,  and  are  fitted  in  the  cylinders  with  great  nicety, 
requiring  none  but  water  packing.  The  parallel  motion  which 
actuates  the  pencil  receives  its  motion  from  the  piston,  a  ball  joint 
connection  being  made  with  the  piston  rod. 

The  upper  side  of  the  piston  has  communication  with  the  atmos- 
phere by  means  of  a  small  hole  in  the  upper  part  of  the  cylinder. 

The  springs  accompanying  an  indicator  are  numbered  to  cor- 
respond to  the  number  of  pounds  of  pressure  per  square  inch  re- 
quired to  cause  a  vertical  movement  of  exactly  one  inch  to  the 
pencil.  For  example:  a  spring  numbered  80  would  require  a 
steam  pressure  of  80  pounds  per  square  inch  to  cause  a  vertical 
movement  of  one  inch  to  the  pencil.  The  tension  of  the  spring  to 
be  used  in  any  particular  case  depends,  of  course,  upon  the  height 
of  the  diagram  desired,  which  in  turn  depends  upon  the  speed  of 
the  engine.  With  a  steam  pressure  of  160  Ibs.  per  square  inch  per 
gauge,  an  80  spring  would  give  a  diagram  2  inches  in  height  above 
the  atmospheric  line.  In  order  to  obtain  satisfactory  diagrams 
from  the  modern  high-speed  engines,  it  has  been  found  necessary 
to  use  springs  of  high  tension  to  limit  the  movement  of  the  piston, 
and  thus  obviate  the  loss  from  friction  that  would  result  from  a 
long  and  rapid  movement.  This  provision  for  a  small  piston  move- 
ment is  all  the  more  necessary  with  the  high  pressures  now  in  use, 
in  order  that  the  pencil  movement  shall  not  be  so  great  as  to  occa- 
sion vibrations  that  will  cause  undulations  in  the  line  of  the  dia- 
gram. 


THE  INDICATOR  AND  ITS  DIAGRAM 


77 


The  height  of  the  diagram  must  be  sufficient  for  practical  use; 
and  to  provide  for  this,  the  parallel  motion  is  designed  to  give  to 
the  pencil  a  motion  of  from  4  to  6  times  that  of  the  piston,  and  it 
is  a  matter  of  the  greatest  importance  that  the  designed  motion  for 
any  instrument  should  not  only  insure  the  straight  line  motion  of 


K 


the  pencil,  but  should  also  maintain  the  fixed  ratio  of  movement 
between  piston  and  pencil,  for  otherwise  the  vertical  movement  of 
one  inch  to  the  pencil  would  not  correspond  in  pounds  pressure  to 
the  number  on  the  spring. 

100.  Figure  21  illustrates  the  construction  of  the  Tabor  indicator, 
which  consists  of  a  steam  cylinder  A,  containing  a  piston  B  and  a 


78  THE  ELEMENTS  OF  STEAM  ENGINEERING 

spring  C.  The  spring  is  secured  to  the  piston  at  one  end  and  to 
the  cover  D  at  the  other  end,  and  the  pressure  of  the  steam  which 
enters  the  indicator  cylinder  through  the  opening  E  compresses  the 
spring  to  a  degree  depending  on  the  pressure.  The  movement  of 
the  piston  is  transferred  to  the  paper  on  the  drum  F,  and  multi- 
plied five  times  by  means  of  the  arrangement  of  levers  shown.  The 
most  noticeable  feature  of  this  indicator  is  the  means  employed  to 
secure  a  straight-line  movement  of  the  pencil.  A  plate  G  contain- 
ing a  curved  slot  is  fixed  in  an  upright  position,  and  a  small  roller 
fixed  to  the  pencil  lever  is  fitted  so  as  to  roll  freely  in  the  slot. 
The  curve  of  the  slot  is  so  formed  that  it  exactly  neutralizes  the 
tendency  which  the  pencil  has  of  describing  a  circular  arc  in  the 
opposite  direction,  and  the  path  of  the  pencil  is  a  straight  line  when 
the  drum  is  not  in  motion.  The  pencil  movement  consists  of  three 
pieces — the  pencil  bar  H,  the  back  link  K,  and  the  piston  rod  link 
L.  The  two  links  K  and  L  are  parallel  to  each  other  in  all  posi- 
tions. The  lower  pivots  of  these  links  and  the  pencil  point  are 
always  in  a  straight  line.  The  paper  drum  is  attached  by  a  cord  8 
to  a  suitable  reducing  motion  from  the  engine;  the  cord  pulls  the 
drum  around  on  its  own  axis  with  a  motion  corresponding  to  that 
of  the  engine  piston,  and  the  return  movement  of  the  drum  is  ob- 
tained by  the  internal  coiled  spring  If. 

The  forward  and  backward  motion  of  rotation  on  its  axis  of  the 
drum  cylinder  that  carries  the  diagram  must  be  exactly  coincident 
with  the  forward  and  backward  motion  of  the  engine  piston,  and 
this  is  readily  obtained  by  a  suitable  connection  by  means  of  cord 
and  pulley  with  the  engine  cross-head,  the  tension  of  the  coiled 
spring  in  the  drum  being  sufficient  to  keep  the  cord  stretched  during 
the  return  stroke  of  the  engine  piston. 

The  length  of  the  diagram  desirable,  as  well  as  the  height,  de- 
pends largely  upon  the  speed  of  the  engine,  slow  speeds  permitting 
longer  and  higher  cards.  The  ordinary  length  is  from  3  to  4  inches, 
so  it  is  obviously  necessary  that  the  motion  of  the  cross-head  be 

Length  of  stroke  of  engine 

reduced  in  the  ratio  — ^ — -pr- — ^ -5-^ — : — ^ —  . 

Length  of  card  desired 

This  arrangement  may  always  be  effected  by  a  linkage  suitable 
to  the  construction  of  the  engine,  care  being  taken  to  avoid  long 
stretches  of  cord,  and  that  the  cord  be  led  from  its  point  of  attach- 
ment to  the  reducing  motion  in  a  direction  parallel  to  the  line  of 
motion  of  the  cross-head,  and  then  over  a  pulley  in  any  direction 


THE  INDICATOR  AND  ITS  DIAGRAM  79 

to  the  indicator;  the  object  being  to  secure  an  equal  motion  of  the 
cord  at  the  drum  and  at  the  point  of  its  'attachment  to  the  reducing 
motion. 

The  "  lazy  tongs  "  and  other  applications  of  the  pantograph  are 
frequently  used  as  reducing  motions,  and  if  carefully  constructed 
the  results  are  accurate. 

101.  The  proper  interpretation  of  indicator  diagrams  reveals  to 
the  engineer  so  many  facts  that  are  essentially  necessary  to  secure 
an  economical  and  efficient  performance  of  a  steam  engine,  that  a 
careful  study  leading  to  a  complete  understanding  of  them  is  a  mat- 
ter of  the  greatest  importance. 

The  principal  value  of  the  indicator  diagram  is  that  it  furnishes 
the  means  of  ascertaining  the  mean  effective  pressure  exerted  on 
the  engine  piston  throughout  the  stroke,  thus  furnishing  the  factor 
which  permits  the  power  of  the  engine  to  be  determined;  but  an 
inspection  of  the  diagram  reveals  information  concerning  particu- 
lars of  the  engine,  of  which  the  following  named  are  the  most  im- 
portant : 

(1)  Whether  the  valves  are  properly  set;  whether  the  admission 
of  steam  is  early  or  late;  whether  the  initial  pressure  is  below  the 
boiler  pressure;  and  the  degree  to  which  the  initial  pressure  is 
maintained  up  to  the  point  of  cut-off. 

(2)  The  point  of  the  stroke  at  which  the  admission  of  steam  to 
the  cylinder  is  cut  off,  and  whether  the  cut-off  is  sharp  or  gradual. 

(3)  The  point  of  the  stroke  at  which  release  takes  place,  and 
the  pressure  of  the  steam  at  that  instant. 

(4)  The  amount  of  back  pressure  opposed  to  the  exhaust,  the 
point  of  the  stroke  at  which  the  exhaust  is  closed,  and  the  amount 
of  compression  at  the  end  of  the  stroke. 

(5)  Whether  the  steam  ports  are  of  adequate  size,  and  whether 
the  valve  or  the  piston  leak. 

(6)  The  approximate  amount  of  steam  consumed  in  a  given 
time,  and  a  number  of  vital  features  concerning  the  balancing  of 
engines. 

Figure  22  is  a  representation  of  the  ideal  indicator  diagram  show- 
ing a  proper  distribution  of  steam  in  the  cylinder,  and  it  is  the 
duty  of  the  engineer  to  set  the  valves  of  his  engine  to  obtain,  as 
nearly  as  possible,  such  a  diagram. 


80 


THE  ELEMENTS  OF  STEAM  ENGINEERING 


102.  The  location  of  the  indicator  and  the  manner  of  its  connec- 
tion with  the  engine  cylinder  is  governed  largely  by  the  construc- 
tion of  the  engine  to  which  it  is  to  be  applied ;  but  wherever  its  loca- 
tion or  the  manner  of  its  connection,  the  principles  governing  its 
action  are  always  the  same.     In  the  case  of  the  ordinary  horizontal 
engine,  the  usual  location  is  midway  in  a  pipe  connecting  the  two 
ends  of  the  cylinder.     A  three-way  cock  in  the  middle  of  the  length 
of  this  pipe  has  a  projecting  nipple  to  which  the  indicator  is  at- 
tached. 

103.  When  about  to  take  a  diagram  the  indicator  should  be 
warmed,  and  the  pipes  blown  through  by  alternately  connecting  the 
indicator  with  the  two  ends  of  the  cylinder  by  means  of  the  three- 
way  cock. 


Fig.  22 


The  length  of  the  drum  string  being  properly  adjusted,  it  is 
connected  with  the  reducing  motion,  and  if  the  pencil  (HHHH) 
be  sharpened  to  a  fine  point,  everything  is  in  readiness  to  take  the 
diagram. 

When  the  three-way  cock  is  closed  to  both  ends  of  the  engine 
cylinder,  a  small  hole  in  the  cock  admits  the  pressure  of  the  atmos- 
phere under  the  indicator  piston,  and  the  piston  being  then  in  a 
state  of  equilibrium  the  Atmospheric  Line,  A  A',  Fig.  22,  may  be 
traced. 

If  now  the  cock  be  turned  so  as  to  admit  steam  from  one  end  of 
the  engine  cylinder  into  the  indicator  cylinder,  the  Admission  Line 
CD  is  instantly  traced,  and  its  height  above  the  atmospheric  line, 
measured  on  the  scale  of  the  spring,  represents  the  initial  pressure 
per  gauge  of  the  steam  admitted  to  the  cylinder. 

The  engine  piston  starting  on  its  stroke,  the  Steam  Line  DEF 


THE  INDICATOR  AND  ITS  DIAGRAM  81 

is  traced  during  the  time  steam  is  being  admitted  into  the  engine 
cylinder,  or  until  cut-off  takes  place. 

F  is  the  point  of  cut-off,  wnere  the  valve  closes  and  prevents 
any  further  admission  of  steam  into  the  cylinder.  The  exact  point 
of  cut-off  is  rather  difficult  to  locate  on  the  diagram,  owing  to  the 
fall  of  steam  pressure  due  to  the  gradual  closing  of  the  port  by  the 
valve,  shown  to  a  small  extent  in  the  fall  in  pressure  from  E  to  F. 

FG  is  the  expansion  curve,  representing  the  fall  in  the  pressure 
of  the  steam  confined  in  the  cylinder  after  cut-off,  due  to  the  ex- 
pansion in  volume  in  forcing  the  piston  to  the  end  of  the  stroke, 

G  is  the  Point  of  Release,  or  the  point  where  the  valve  opens  to 
exhaust,  thereby  releasing  the  steam  from  the  cylinder. 

GH  is  the  Exhaust  Line,,  which  is  traced  in  the  interval  between 
release  and  the  end  of  the  stroke,  the  pressure  falling  rapidly  to 
that  of  the  back  pressure  opposed  to  the  exhaust. 

HI  is  the  Back  Pressure  Line,  and  indicates  the  pressure  oppos- 
ing the  piston  on  the  return  stroke.  In  non-condensing  engines 
this  line  coincides  with,  or  is  slightly  above,  the  atmospheric  line, 
and  in  condensing  engines  it  is  below  the  atmospheric  line  a  dis- 
tance corresponding  to  the  vacuum  obtained;  but  in  either  case  it 
is  back  pressure.  Vacuum  is  expressed  in  inches  of  mercury,  and 
since  one  cubic  inch  of  mercury  weighs  0.491  pound,  the  inches  of 
vacuum  multiplied  by  0.491  will  give  the  pressure  equivalent  to  the 
vacuum. 

I  is  the  point  of  exhaust  closure,  where  compression  begins. 
This  point,  like  those  of  cut-off  and  release,  is  difficult  to  locate 
exactly. 

1C  is  the  Compression  Curve,  and  represents  the  rise  in  pressure 
due  to  the  piston  compressing  the  steam  remaining  in  the  cylinder 
after  the  exhaust  has  closed. 

For  the  study  of  the  diagrams  and  for  computations  involving 
pressures,  it  is  necessary  to  locate  the  Vacuum  Line  00',  or  line  of 
no  pressure,  from  which  all  pressures  should  be  measured  in  order 
that  they  may  be  absolute.  The  location  of  this  line  is  parallel  to 
the  atmospheric  line  and  at  a  distance  from  it  equal  to  the  pressure 
of  the  atmosphere,  the  average  value  of  which  is  14.7  Ibs.,  measured 
on  the  scale  of  the  indicator  spring. 

Of  equal  importance  is  the  establishment  of  the  Clearance  Line 
OB,  perpendicular  to  the  atmospheric  line,  and  at  a  distance  from 
the  end  of  the  diagram  equal  to  the  same  percentage  of  the  length 
6 


82  THE  ELEMENTS  OF  STEAM  ENGINEERING 

of  the  stroke  that  the  volume  of  the  clearance  space  of  the  cylinder 
bears  to  the  piston  displacement. 

104.  Clearance,  as  we  have  already  seen,  is  the  volume  of  the 
space  between  the  valve  face  and  the  engine  piston  when  the  piston 
is  at  the  end  of  its  stroke,  and  this  clearance  has  to  be  filled  with 
steam  at  each  end  of  the  cylinder  for  every  revolution  of  the  engine. 
This  steam  must  come  from  the  boiler,  or  from  the  steam  left  in 
the  cylinder  by  an  early  exhaust  closure,  or  from  both.  The  amount 
of  clearance  varies  in  different  styles  of  engines.  In  engines  of 
slow  speed  and  long  stroke,  the  variation  is  from  2  to  5  per  cent; 
for  high-speed  engines  with  short  stroke,  it  may  be  as  much  as  10 
per  cent;  while  for  marine  engines,  a  clearance  of  20  per  cent  is 
not  uncommon. 

Engine  builders  usually  measure  the  clearance  volume  very  care- 
fully, but  in  the  absence  of  any  data  on  the  subject,  an  approxima- 
tion of  the  clearance  of  an  engine  may  be  determined  graphically 
from  the  expansion  or  from  the  compression  curves  of  a  well-de- 
fined indicator  diagram.  , 

Thus  in  Fig.  22,  select  two  points  &  and  c  in  the  compression 
curve  as  far  apart  as  possible,  and  through  them  draw  an  indefinite 
line  intersecting  the  vacuum  line  at  a.  From  c  lay  off  cd  =  ba, 
and  through  d  draw  OB  perpendicular  to  00'.  OB  will  be  the 
clearance  line,  and  BD  will  represent  the  volume  of  the  clearance 
to  the  same  scale  that  AA'  represents  the  volume  of  the  cylinder. 
This  construction,  as  well  as  the  two  that  follow,  assumes  the  curves 
of  expansion  and  compression  to  be  hyperbolic,  the  asymptotes  of 
which  are  the  vacuum  and  clearance  lines.  These  assumptions  are 
sufficiently  accurate  for  practical  purposes,  and  as  they  greatly 
simplify  calculations,  they  are  always  made. 

The  method  just  given  of  locating  the  clearance  line  depends 
upon  the  property  of  the  hyperbola  that,  if  any  secant  line  be  drawn, 
the  portions  of  the  secant  intercepted  between  the  curve  and  its 
asymptotes  are  equal. 

Two  other  constructions  are:  (a)  The  line  &c  joining  the  two 
selected  points  on  the  curve  is  a  diagonal  of  a  rectangle  constructed 
on  the  curve,  the  two  sides  of  which  are  parallel  to  the  vacuum  line ; 
if  now  a  diagonal  be  drawn  through  the  outer  corners  of  the 
rectangle,  and  produced  until  it  intersects  the  vacuum  line,  the 
point  of  intersection  0  will  be  the  center  of  the  curve,  or  the  inter- 


THE  INDICATOR  AND  ITS  DIAGRAM  83 

section  of  the  asymptotes;  hence  OB,  perpendicular  to  00',  is  the 
clearance  line.  (&)  If  two  points  V  and  c'  be  taken  on  the  expan- 
sion curve,  the  construction  is  the  same. 

Though  these  constructions  are  mathematically  correct  when  the 
curves  are  hyperbolas,  they  give  results  by  no  means  reliable  when 
applied  to  the  curves  of  actual  diagrams. 

Clearance  in  an  engine  occasions  a  loss  when  the  consumption 
of  steam  per  unit  power  is  considered,  but  there  are  practical  con- 
siderations which  render  its  existence  highly  desirable,  if  not  neces- 
sary. The  clearance  space  between  the  piston  and  the  cylinder 
head  when  the  piston  is  at  the  end  of  its  stroke,  gives  space  for  the 
variable  amount  of  water  which  is  always  present  in  a  cylinder  and 
doubtless  prevents  serious  accidents  which  might  otherwise  occur. 

Since  the  piston  does  not  traverse  the  clearance  space,  the  clear- 
ance steam  performs  no  initial  work;  that  is,  it  does  no  work  during 
the  period  of  admission,  but  after  cut-off  its  effect  is  to  raise  the 
pressure  during  expansion,  and  thus  increase  the  area  of  the  expan- 
sion part  of  the  diagram.  If  there  were  neither  expansion  nor 
compression,  the  clearance  steam  would  perform  no  work  and 
would  be  a  total  loss  in  the  exhaust.  On  the  other  hand,  if  the  ex- 
pansion curve  were  carried  down  to  the  back  pressure,  and  the  com- 
pression curve  carried  up  to  the  initial  pressure,  there  would  be 
absolutely  no  loss  from  clearance. 

These  latter  conditions  are  rarely,  if  ever,  realized  in  practice, 
therefore  there  is  always  -a  loss  from  clearance,  and  this  loss  is 
greater  as  the  clearance  is  proportionally  large. 

One  effect  of  cushioning  is  to  reduce  the  loss  from  waste  of  steam 
in  the  clearance  space ;  but  its  most  important  effect  is  that  it  pro- 
vides for  smooth  running  of  the  engine  by  preventing  shocks  at  the 
end  of  the  stroke.  It  is  especially  desirable  that  the  diagram  of  a 
high-speed  single-cylinder  engine  should  have  its  compression  corner 
well  rounded. 

105.  The  influence  of  clearance  on  the  ratio  of  expansion  is  so 
marked  that  it  is  essential  that  the  amount  of  the  clearance  be 
known  and  allowed  for. 

In  any  case  the  ratio  of  expansion  is  the  volume  of  steam,  includ- 
ing clearance,  at  the  end  of  the  stroke,  divided  by  the  volume  of 
steam,  including  clearance,  up  to  the  point  of  cut-off.  Denoting 
the  ratio  of  expansion  by  r,  the  initial  volume,  or  volume  up  to 


84  THE  ELEMENTS  OF  STEAM  ENGINEERING 

cut-off,  including  clearance,  by  viy  and  the  final  volume,  including 
clearance,  by  v2,  we  shall  have  in  all  cases  : 


Keferring  to  Fig.  22,  the  ratio  of  expansion,  if  clearance  were 

PQI 

neglected,  would  be          '     ^  clearance  were  considered,  the  vol- 


ume of  steam  that  expanded  after  cut-off  at  F  would  be  OP  +  PK, 
and  the  final  volume  would  be  OP  +  PO'  ;  therefore  the  real  ratio 

OP  +  PO' 
of  expansion  would  be  /i      4    PAT*     Since  ^ne  cross-sectional  area 


of  the  cylinder  is  uniform,  the  volume  displaced  by  the  piston  at 
any  point  is  directly  proportional  to  the  fraction  of  the  stroke 
passed  through  up  to  that  point,  and  volumes,  therefore,  may  be 
represented  by  the  corresponding  fractions  of  stroke.  In  like  man- 
ner the  clearance  volume,  when  divided  by  the  cross-sectional  area 
of  the  cylinder,  will  be  expressed  in  some  fractional  part  of  the 
stroke. 

If,  therefore,  we  denote  the  full  stroke  of  the  piston  by  unity,  it 
may  also  represent  the  volume  displaced  by  the  piston  in  one 
stroke,  in  which  case  the  fraction  of  the  stroke  denoting  the  cut-off 
will  represent  the  volume  displaced  by  the  piston  up  to  the  point  of 
cut-off.  If  the  fraction  of  the  stroke  performed  by  the  piston  up 
to  the  point  of  cut-off  be  denoted  by  k,  and  if  c  be  the  fractional 
part  of  the  stroke  representing  the  clearance,  we  shall  have  Tc  -j-  c 
for  our  initial  volume,  and  1  +  c  for  our  final  volume.  The  ratio 

of  expansion  will  then  be  r  =.  —  =  --,          • 

To  illustrate  the  effect  of  clearance  on  the  ratio  of  expansion, 
and  to  show  the  application  of  the  formula  just  given,  we  will  con- 
sider a  numerical  example. 

EXAMPLE  I.  —  Diameter  of  cylinder,  42";    stroke,  4';    cut-off, 
£  stroke  ;  clearance,  8  per  cent.     Find  the  ratio  of  expansion. 
Here  we  have  : 

Stroke  volume  displaced  by  piston  =  -      T^X  144  -  =  38.5  cu.  ft. 

Volume  of  clearance  =  38.5  X  0.08  =  3.08  cu.  ft. 
V2  =  38.5  +  3.08  =  41.58  cu.  ft. 


THE  INDICATOR  AND  ITS  DIAGRAM  85 

oo  t 

Volume  displaced  by  piston  up  to  cut-off  =  -^—  =  9.625  cu.  ft. 
v1  =  9.625  +  3.08  =  12.705  cu.  ft. 


Had  the  clearance  been  neglected,  we  should  have  had  r  = 

38  5 
g-g^g  =  4  ,  a  result  which  would  affect  the  calculated  mean  pres- 

sure. 

The  work  would  have  been  much  simplified  by  the  use  of  the 

1  4-  c 

formula    r  =  j—   -  ;     for,    by    substitution,,    we    would    have 
K  +  c 

as  before-  • 


106.  The  question  of  expansion  in  the  compound,  triple,  or  any 
other  stage  expansion  system,  is  not  different  from  that  in  the 
single-cylinder  engine,  so  far  as  the  total  rate  of  expansion  is  con- 
cerned. To  know  the  volumes  of  the  high-  and  low-pressure  cylin- 
ders and  their  clearances,  together  with  the  point  of  cut-off  in  the 
high-pressure  cylinder  is  all  that  is  necessary  to  know  in  order  to 
determine  the  total  rate  of  expansion.  The  intervening  cylinders, 
receiver  spaces,  and  point  of  cut-off  in  the  low-pressure  cylinder 
have  nothing  to  do  with  the  question. 

As  in  the  simple  engine,  the  total  ratio  of  expansion  in  any  stage 
expansion  engine  is  the  ratio  of  the  final  volume  occupied  by  the 

steam  to  the  initial  volume,  or  r  =  —  ,  in  which  v2  is  the  volume 

of  the  low-pressure  cylinder  and  its  clearance.  v±  is  the  initial  vol- 
ume, or  the  volume  up  to  cut-off,  including  clearance,  of  the  high- 
pressure  cylinder. 

Neglecting,  for  the  present,  the  question  of  compression  and 
cylinder  condensation,  the  initial  volume  of  steam  must  eventually 
fill  the  whole  of  the  low-pressure  cylinder  and  its  clearance  space 
at  one  end;  therefore,  to  obtain  the  total  ratio  of  expansion,  we 
have  simply  to  divide  this  final  volume  v2  by  the  initial  volume  v^. 
This  may  be  better  understood  if  illustrated  by  a  numerical  example. 

EXAMPLE  II.  —  A  two-cylinder  compound  engine  has  a  high-pres- 
sure cylinder  26"  in  diameter,  a  low-pressure  cylinder  50"  in  diam- 
eter, and  a  stroke  of  30".  Initial  absolute  pressure  of  steam,  115  Ibs. 
per  sq.  inch.  Cut-off  in  high-pressure  cylinder  at  0.5  stroke,  and 


86  THE  ELEMENTS  OF  STEAM  ENGINEERING 

in  low-pressure  cylinder  at  0.8  stroke.     Clearance  of  each  cylinder 
5  per  cent.     It  is  required  to  find  the  total  ratio  of  expansion. 

The  area  of  a  26"  piston  is  530.93  sq.  inches,  and  of  a  50"  piston, 
1963.5  sq.  inches.  The  clearance  of  5  per  cent  is  equivalent  to  the 
addition  of  30  X  0.05  =  1.5  inches  to  the  stroke.  Neglecting  com- 
pression >and  cylinder  condensation,  we  shall  have 

530.93  X  16.5  _  -  07 
1728 

cubic  feet  of  steam  supplied  the  high-pressure  cylinder  per  stroke, 
which  is  the  initial  volume  vx. 

All  of  this  steam  must  finally  occupy  the  low-pressure  cylinder 
and  its  clearance  at  one  end,  hence  its  final  volume  v2  will  be 

1963  5  X  31  5 

'-iiyou —  —  =  35.793  cu.  ft.       The  ratio  of  expansion  is  then 

r  =  !!?  =  35;793  = 
vl        5.07 

No  account  has  been  taken  of  the  volume  of  the  high-pressure 
cylinder  filled  with  steam  after  cut-off,  nor  of  that  of  the  clearance 
space  between  the  cylinders,  nor  has  the  cut-off  in  the  low-pressure 
cylinder  been  considered,  for  the  reason  that  they  do  not  enter  into 
the  question  at  all. 

The  effect  of  the  receiver  is  to  make  the  initial  pressure  lower  in 
the  low-pressure  cylinder  than  it  would  have  been  had  the  exhaust 
been  direct  from  the  high-pressure  to  the  low-pressure  cylinder, 
and  this  reduction  is  due  to  the  unrestricted  expansion  of  the  steam 
when  it  enters  the  receiver  space.  The  receiver  only  plays  the  part 
of  a  large  clearance  space. 

A  low-pressure  cut-off  will  increase  the  receiver  pressure  and 
therefore  the  initial  pressure  in  the  low-pressure  cylinder,  with  a 
consequent  increase  in  power  of  the  low-pressure  engine.  The  in- 
crease in  receiver  pressure  is  an  increase  in  the  back  pressure  on 
the  high-pressure  piston,  and  occasions  a  decrease  in  the  work  of 
the  high-pressure  engine. 

So  it  is  seen  that  the  use  of  a  low-pressure  cut-off  is  to  equalize 
the  power  of  the  two  engines,  and  has  nothing  to  do  with  the  total 
rate  of  expansion. 

This  may  be  made  more  clear  by  considering  the  weight  of  steam 
used  per  stroke  in  *bhe  above  example. 

The  specific  volume  of  steam  at  115  Ibs.  pressure  is  3.839  cu.  feet, 


THE  INDICATOR  AND  ITS  DIAGRAM  87 

5  07 
consequently  »  OOQ  =  1.3207  Ibs.  steam  enters  the  high-pressure 

cylinder  per  stroke.  At  the  commencement  of  the  return  stroke  of 
the  high-pressure  piston  the  delivery  of  this  steam  into  the  receiver 
will  begin,  and  it  will  have  been  delivered  when  the  return  stroke  is 
completed.  Its  withdrawal  from  the  receiver  will  begin  with  the 
commencement  of  the  stroke  of  the  low-pressure  piston,  and  must 
all  be  drawn  out  by  the  time  the  low-pressure  cut-off  takes  place. 
If  it  were  not  all  drawn  out,  the  pressure  in  the  receiver  would  in- 
crease, as  the  engine  ran,  until  no  exhaust  from  the  high-pressure 
cylinder  could  take  place.  If  more  were  drawn  off,  a  point  would 
soon  be  reached  at  which  a  vacuum  would  exist  in  the  receiver.  As 
both  these  assumptions  are  absurdly  impossible,  it  follows  that  the 
receiver  has  nothing  to  do  with  the  total  ratio  of  expansion. 

The  accuracy  of  our  conclusions  may  be  checked  by  a  further 
consideration  of  the  numerical  example. 

We  have  found  that  1.3&07  Ibs.  steam  are  used  per  stroke,  and 
that  the  total  ratio  of  expansion  is  7.06.  If  these  figures  are  cor- 
rect, the  calculated  volume  of  steam  found  in  the  low-pressure  cyl- 
inder and  clearance  at  the  end  of  the  stroke  should  be  equal  to  the 
combined  volume  of  the  low-pressure  cylinder  and  clearance. 

The  relation  between  the  pressure  and  volume  of  saturated  steam 
is  pv&  —C:,  therefore,  ptv^  =p2v^*,  in  which  p^  is  the  initial 
pressure  and  v±  the  initial  volume  in  the  high-pressure  cylinder, 
and  p2  the  final  pressure  and  v2  the  final  volume  in  the  low-pressure 
cylinder.  We  have: 


7.06  log  0.84880    Hog  9.92881  —  10 
17     log  1.23045 
16  colog  8.79588  —  10 

(7.06)**  log  0.90186     Hog  9.95514  —  10 
115  log  2.06070 

p2  —  14.417  log  1.15884 

The  volume  of  one  pound  of  steam  at  pressure  of  14.417  Ibs.  per 
sq.  inch  is  26.82  cu.  ft.,  and  since  there  are  1.3207  Ibs.,  the  total 
volume  it  occupies  is  1.3207  X  26.82  =  35.42  cu.  ft.  Allowing  for 
abbreviated  decimals,  this  corresponds  with  the  calculated  volume 
of  the  low-pressure  cylinder  and  clearance,  as  it  should. 


THE  ELEMENTS  OF  STEAM  ENGINEERING 

To  apply  the  formula,  r  =  ^  *  ^,  in  finding  the  total  ratio  of 

expansion  in  a  stage  expansion  engine,  we  must  substitute  for  unity 
in  the  numerator  the  volumetric  ratio  of  the  low-pressure  cylinder  to 
the  high-pressure  cylinder,  and  for  c  the  clearance  percentage  of  this 
ratio.  Denoting  the  ratio  of  the  low-pressure  cylinder  to  the  high- 
pressure  cylinder  by  ^,  we  would  then  have: 


In  the  above  example  f  ==  3.698,  hence  r= 


., 

(/CD)  0.5  +  0.05 

=  7.06,  as  before. 

In  our  calculations  the  effect  of  compression  has  been  neglected, 
but  the  only  way  this  could  affect  the  question  would  be  to  reduce 
the  quantity  of  steam  withdrawn  from  the  boiler  at  each  stroke  and 
this  may  be  regarded  as  virtually  increasing  slightly  the  ratio  of 
expansion,  because  a  less  weight  of  fresh  steam  would  be  used  each 
stroke. 

The  effect  of  cylinder  condensation  has  also  been  neglected,  but 
this,  too,  would  occasion  a  virtual  augmentation  of  the  ratio  of  ex- 
pansion, because  a  smaller  weight  of  steam  than  that  delivered  to 
the  high-pressure  cylinder  would  be  found  in  the  low-pressure  cylin- 
der at  the  end  of  the  stroke. 

In  either  case,  the  steam  must  finally  fill  the  low-pressure 
cylinder  and  clearance  whatever  the  weight. 

The  effect  of  cutting  off  in  the  low-pressure  cylinder  is  to  main- 
tain the  average  pressure  in  the  receiver  higher  than  it  would  be  if 
there  were  no  cutting  off  of  the  steam,  -and  therefore  to  reduce  the 
power  developed  in  the  high-pressure  cylinder  by  increasing  its  back 
pressure.  Whether  the  steam  is,  or  is  not,  cut  off  in  the  low-pres- 
sure cylinder,  the  same  weight  of  steam  must  find  its  way  into  that 
cylinder  at  each  stroke  ;  and  if,  by  means  of  the  cut-off,  a  less  space 
be  allowed  for  the  reception  of  the  steam,  the  pressure  will  increase 
accordingly. 

107.  Wire-drawing.  —  Wire-drawing  is  a  technical  name  for  the 
reduction  in  pressure  of  steam  resulting  from  its  passage  through 
contracted  areas,  and  its  effect  is  to  reduce  the  efficiency  of  the 
steam. 

The  initial  pressure  on  the  piston  is  always  appreciably  less  than 


THE  INDICATOR  AND  ITS  DIAGRAM  89 

the  boiler  pressure,  and  this  difference  is  to  be  attributed  to  wire- 
drawing in  the  steam  pipe.  The  gradual  falling  of  the  steam  line, 
as  the  piston  advances  to  the  point  of  cut-off,  is  wire-drawing  due 
to  the  ports  being  of  insufficient  area  to  admit  steam  fast  enough  to 
maintain  a  full  pressure  behind  the  piston,  and  also  to  the  action 
of  the  slide  valve  in  closing  the  port  gradually.  This  movement  of 
the  slide  valve  prevents  a  sharp  cut-off,  and  the  exact  point  at  which 
the  port  closes  is,  therefore,  difficult  to  locate  on  the  diagram.  To 
make  the  ports,  and  therefore  the  valve,  sufficiently  large  to  avoid 
wire-drawing  would  be  a  matter  of  more  serious  consequence  than 
the  defect  itself. 

108.  In  order  that  the  exhaust  steam  may  flow  from  the  cylinder 
of  a  condensing  engine  to  the  condenser,  or  into  the  atmosphere 
from  the  cylinder  of  a  non-condensing  engine,  the  actual  back  pres- 
sure must  be  greater  than  the  pressure  in  the  condenser  in  the  one 
case,  or  greater  than  the  atmospheric  pressure  in  the  other,  and  this 
excess  of  pressure  depends  largely  upon  the  freedom  of  passage  for 
the  exhaust  steam  from  the  cylinder  to  the  condenser  or  to  the 
atmosphere. 

The  release  of  the  steam  at  from  -^  to  y1^  of  the  stroke  from  the 
end  assists  materially  in  the  freedom  of  the  exhaust;  this  is  neces- 
sary with  a  condensing  engine  to  insure  a  nearly  complete  vacuum 
when  the  piston  starts  on  its  return  stroke,  and  with  a  non-con- 
densing engine  it  enables  the  exhaust  steam  to  begin  its  flow  into 
the  atmosphere  before  the  return  stroke  commences. 

In  practice  the  exhaust  is  closed  at  some  point  E,  Fig.  23,  before 
the  piston  arrives  at  the  end  of  its  stroke,  and  the  steam  entrapped 
in  the  cylinder  is  compressed  by  the  advancing  piston,  its  pressure 
rising  to  some  point  F,  when  the  valve  opens  to  lead,  the  pressure 
rising  suddenly  to  A,  and  a  new  stroke  commences. 

109.  Figures  23  and  24  represent  diagrams  from  a  non-con- 
densing and  from  a  condensing  engine,  respectively. 

The  diagrams  taken  from  an  engine  of  proper  design  and  adjust- 
ment do  not  differ  materially  from  the  theoretical  diagram,  but  it 
requires  careful  study  and  discriminating  judgment  to  make  proper 
use  of  the  information  presented  by  them,  a  fact  which  may  be  ap- 
preciated when  it  is  considered  that  the  only  'absolute  information 
a  diagram  gives  is  the  varying  pressure  of  the  steam  in  the  cylinder. 


90 


THE  ELEMENTS  OF  STEAM  ENGINEERING 


The  full  line  diagram  of  Fig.  25  would  indicate  a  very  satisfac- 
tory performance  of  the  engine  from  which  it  was  taken.  The 
dotted  lines  illustrate  some  possible  defects  of  an  engine  which 
would  readily  be  detected  by  the  indicator.  Thus,  the  fall  of  the 
steam  line  to  rib  would  indicate  wire-drawing;  the  line  cd  would 


Perfect 


show  that  the  "  release  "  was  too  early,  and  the  line  ef  that  it  was 
too-  late;  the  inclining  of  the  admission  line  to  the  left,  as  at  ga, 
would  show  the  lead  to  be  too  great,  and  its  inclination  to  the  right, 
as  at  hi,  would  show  insufficient  lead. 


24 


110.  Mean  Effective  Pressure  from  Diagram. — There  are  two 
methods  of  obtaining  the  mean  effective  pressure  from  a  diagram : 
(a)  By  the  use  of  the  planimeter.  (&)  By  the  measurement  of 
ordinates. 

The  planimeter  'is  an  ingenious  instrument  by  means  of  which 
the  mean  pressure  may  be  obtained  accurately  and  more  quickly 


THE  INDICATOR  AND  ITS  DIAGRAM 


91 


than  by  the  method  of  ordinates,  though  the  latter  method  is  the 
more  common. 

To  obtain  the  mean  effective  pressure  by  the  method  of  ordinates, 
erect  perpendiculars  to  the  atmospheric  line  touching  the  extreme 
ends  of  the  diagrams.  Divide  the  space  between  these  perpendicu- 
lars into  ten  equal  parts  and  erect  perpendiculars  to  the  atmospheric 


line  at  the  middle  points  of  these  divisions.  The  first  and  last  of 
these  new  divisions  will  be  ^V  the  length  of  the  diagram  from  the 
ends,  and  the  common  interval  between  them  -will  be  TV  of  the 
length  of  the  diagram.  One-tenth  the  sum  of  the  lengths  of  the 
ordinates  will  be  the  length  of  the  mean  ordinate,  or  the  mean  effec- 
tive pressure  in  pounds  per  square  inch  on  the  piston,  measured  to 
the  scale  of  the  indicator  spring. 


Sccr/e  /OO. 


me.p.  - 


Jnch. 


The  diagrams  of  Fig.  26  were  taken  from  the  high-speed  engine 
in  the  mechanical  laboratory  of  the  Baltimore  Polytechnic  Insti- 
tute. The  sum  of  the  lengths  of  the  ordinates  of  the  diagrams  from 
the  head  and  crank  ends  of  the  cylinder  are  3.2  and  3.25  inches 
respectively,  and  the  scale  of  the  spring  is  100  pounds  to  the  inch. 
The  mean  effective  pressure  is,  therefore,  for  one  revolution, 


92  THE  ELEMENTS  OF  STEAM  ENGINEERING 

The  diagrams  from  the  two  ends  of  the  cylinder  should  be  taken 
simultaneously  if  two  indicators  are  used,  or  one  immediately  after 
the  other  if  only  one  be  used. 

111.  Having  found  from  the  diagram  the  mean  pressure  in 
pounds  per  square  inch  acting  on  the  piston  throughout  one  revolu- 
tion, the  mean  total  pressure  in  pounds  will  be  the  product  of  this 
mean  pressure  and  the  area  of  the  piston  in  square  inches ;  and  the 
work  per  revolution  in  foot-pounds  will  be  found  by  multiplying 
this  pressure  by  twice  the  length  of  the  stroke  in  feet ;  and  the  work 
per  minute  will  be  found  by  multiplying  the  work  per  revolution  by 
the  number  of  revolutions  per  minute.  This  last  product  divided 
by  33,000  will  express  the  work  in  horse-power. 

The  mean  pressure  having  been  found  from  the  indicator  dia- 
gram, the  horse-power  is  called  Indicated  Horse-Power,  usually 
written  I.  H.  P.,  and  expresses  the  gross  work  done  on  the  piston 
by  the  steam.  It  is  usual  to  express  the  indicated  horse-power  by 
an  easily  remembered  formula,  as  follows : 

y  TT  p        pLaN 
~  33,000' 

in  which  p  =  mean  effective  pressure  in  pounds  per  sq.  inch  on  the 

piston. 

L  =  length  of  stroke  in  feet. 
a  =  area  of  piston  in  square  inches. 
N  =  number  of  strokes  per  minute  =  twice  the  number 

of  revolutions  per  minute. 

It  will  be  observed  that  the  formula  for  I.  H.  P.  is  an  algebraic 
expression  for  the  definition  that  work  is  "  pressure  into  its  path," 
or  "  force  into  its  distance." 

For  p  X  &  is  equal  to  the  pressure,  or  force,  in  pounds ;  L  X  N 
is  the  path,  or  distance,  in  feet  moved  through  in  a  minute;  there- 
fore, the  product  (p  X  a)  (L  X  N)  is  given  in  foot-pounds  per 
minute;  and  as  33,000  foot-pounds  of  work  per  minute  is  the 

measure  of  one  horse-power  it  becomes  evident  that  |^-QQQ  is  the 
measure  of  the  work  of  an  engine  in  terms  of  I.  H.  P. 

The  I.  H.  P.  formula  also  expresses  that  the  external  work  of 
steam  is  equal  to  PV,  in  which  P  is  the  pressure  per  square  foot 


THE  INDICATOR  AND  ITS  DIAGRAM  93 

and  V  the  volume  in  cubic  feet  displaced  by  the  piston  in  a  minute. 


_    PV 
33,000  33,000  =  33,000  " 

Only  the  net  area  of  the  piston  must  be  used  in  calculating  the 
horse-power  of  an  engine  ;  and  as  the  steam  on  one  side  of  the  piston 
acts  only  on  an  area  equal  to  that  of  the  piston  diminished  by  the 
cross-sectional  area  of  the  piston  rod,  this  latter  area  must  be  de- 
ducted from  that  side. 

If  ar  denotes  the  cross-sectional  area  of  the  piston  rod,  then  the 
mean  net  area  of  piston  to  be  used  will  be 


In  the  stage  expansion  engine,  the  I.  H.  P.  for  each  cylinder  is 
obtained,  and  their  sum  is  the  total  power  of  the  engine. 


Should  a  diagram  be  looped  as  in  Fig.  27,  the  area  adc  represents 
negative  work,  and  in  obtaining  the  mean  pressure  from  such  a 
diagram,  the  lengths  of  the  ordinates  included  in  the  loop  should 
be  subtracted  from  the  total  length  of  those  in  the  area  abe. 

A  loop  like  that  in  Fig.  27  would  be  occasioned  by  excessive 
expansion.  At  the  point  a  where  the  expansion  curve  crosses  the 
back  pressure  line,  it  is  evident  that  the  pressures  on  each  side  of  the 
piston  are  equal,  and  a  cut-off  which  would  occasion  an  expansion 
so  excessive  as  to  reduce  the  steam  pressure  to  a  point  below  the 
back  pressure  opposed  to  the  piston  would  manifestly  be  too  early. 
The  theoretical  limit  of  expansion  is  such  that  the  terminal  pressure 
should  be  just  equal  to  the  back  pressure.  In  practice  the  terminal 
pressure  is  in  excess  of  this,  varying  from  24  to  28  pounds  in  non- 
condensing  engines,  and  from  9  to  12  pounds  in  condensing  engines. 


94  THE  ELEMENTS  OF  STEAM  ENGINEEKING 

In  actual  practice,  a  loop  in  a  diagram  would  very  likely  indicate 
that  the  engine  was  underloaded. 

112.  It  has  already  been  shown  that  mechanical  work  is  produced 
by  a  force  working  through  a  distance.  In  the  case  of  any  gas 
working  within  a  cylinder  against  a  piston,  the  force  will  be  the 
mean  value  of  the  pressure  of  the  gas  multiplied  by  the  area  of  the 
piston,  and  the  distance  will  be  the  stroke  of  the  piston.  In  order 
that  the  work  may  be  expressed  in  foot-pounds,  the  force  must  be 
expressed  in  pounds  and  the  distance  in  feet. 

It  is  seen,  then,  that  the  area  of  an  indicator  diagram  is  the 
measure  of  the  work  performed  in  a  cylinder;  for  its  area  is  the 
product  of  its  mean  ordinate  and  its  length,  factors  representing  the 
mean  effective  pressure  on  the  piston  in  pounds  per  square  inch  and 
the  distance  moved  through  in  feet  respectively. 

PROBLEMS, 

27.  Diameter  of  H.  P.  cylinder,  15";  diameter  of  L.  P.  cylinder, 
30";  cut-off  in  H.  P.  cylinder,  0.25  stroke;  clearance  of  each  cylin- 
der, 9  per  cent:   find  total  ratio  of  expansion.  Ans.  12.82. 

28.  Diameter  of  H.  P.  cylinder,  42",- and  of  L.  P.  cylinder,  78"; 
cut-off  in  H.  P.  cylinder,  0.3  stroke;  clearance  of  H.  P.  cylinder, 
10  per  cent,  and  of  L.  P.  cylinder  12  per  cent:  find  total  ratio  of 
expansion.  Ans.  9.6575. 

29.  Eatio  of  expansion,  3.2727;    cut-off,  0.25  stroke:    find  the 
clearance.  Ans.  8  per  cent. 

30.  Eatio  of  expansion,  3.2727;  clearance,  8  per  cent:   find  the 
point  of  cut-off.  Ans.  0.25. 

31.  Diameter  of  cylinder,  9";   stroke,  10";   cut-off,  0.28  stroke; 
clearance,  6.5  per  cent.     Initial  pressure  of  steam,  95  Ibs.  per  square 
inch  absolute,  the  specific  volume  of  which  is  4.57  cubic  feet.     Find 
ratio  of  expansion  and  terminal  pressure,  assuming  the  steam  used 
to  be  saturated.     Prove  the  accuracy  of  the  results  by  showing  that 
the  calculated  volume  of  the  steam  in  the  cylinder  and  clearance  at 
the  end  of  the  stroke  is  equal  to  the  combined  volume  of  the  cylinder 
and  clearance.     The  specific  volume  of  steam  at  the  terminal  pres- 
sure is  14.04  cubic  feet.  Ans.  r  =  3.087.     p2  =  28.68. 


THE  INDICATOR  AND  ITS  DIAGRAM  95 

32.  A  two-cylinder  compound  engine,  the  dimensions  of  which 
are  15",  30",  X  18",  has  <a  clearance  of  5  per  cent  in  each  cylinder. 
The  initial  absolute  pressure  is  100  Ibs.,  and  the  cut-off  is  at  f  of 
the  stroke.  Find  the  total  ratio  of  expansion  and  the  terminal  pres- 
sure in  the  low-pressure  cylinder.  Prove  the  accuracy  of  the  work 
by  showing  that  the  cubic  feet  of  steam  in  the  low-pressure  cylinder 
and  clearance  at  the  end  of  the  stroke  equals  the  volume  of  the  low- 
pressure  cylinder  and  its  clearance.  The  steam  is  assumed  to  be 
saturated,  and  the  specific  volume  may  be  found  by  the  formula 
=  482.  Ans.  r  =  9.882.  p2  =  8.768. 


CHAPTER  IX. 

MEAN  PRESSURE  OF  A  GAS  EXPANDING  WITHIN  A 
CYLINDER, 

113.  With  the  understanding  of  work  and  of  the  indicator  dia- 
gram obtained  from  Chapter  VIII,  it  will  now  be  shown  how  the 
mean  pressure  of  any  gas  working  within  a  cylinder  may  be  calcu- 
lated, the  gas  expanding  according  to  any  law. 

Referring  to  Fig.  28, 


L 


.  28 


Pi  =  initial  pressure  in  pounds  per  square  inch. 

p2  =  terminal  pressure. 

p3  =  back  pressure. 

pm  —  mean  absolute  pressure. 

pe  =  mean  effective  pressure  =  pm  —  py 

v^  =  the  initial  volume,  and  v2  =  the  final  volume,  of  the  gas. 

r  —  -^ *  =  ratio  of  expansion,  and  v2  =  rv^. 

The  area  of  the  diagram  represents  the  gross  work  performed  by 
the  volume  of  gas  v±  of  initial  pressure  p±  in  driving  a  piston  in  a 
cylinder  through  a  stroke  AE  against  a  back  pressure  pa,  the  vol- 
ume v±  of  pressure  p^  expanding  to  the  volume  v2  and  pressure  p2. 


MEAN  PRESSURE  OF  GAS  EXPANDING  WITHIN  CYLINDER      97 

If  pm  denotes  the  mean  absolute  pressure  of  the  gas,  then 

Gross  Work  =  pmv2 
whence 

_  Gross  Work Area  of  Diagram 

The  area  of  the  rectangle  ABCF  is  p^v^,  and  if  the  equation  of  the 
expansion  curve  be  pvn  =  C,  a  constant,  then  the  expansion  area 


FCDE  is    I  pdv. 

1)  V  n 

But  pvn  —  p^v^  =  p2v2n,  whence  p  —  —r  • 

Substituting  this  value  of  p  in  the  integral  expression  we  get 

Expansion  area  FCDE  of  diagram  =    /     -—n —  • 

Jvi       v 

Therefore, 

Whole  area  of  diagram  —  plvl  -f  pp*  I   2  ~ 

1  1  J  ^ 

/»rUl 

—  Pivi  +  P\v*  I  v~"dv,  since  v.2  —  rvl. 

*/Vl 

By  integration  we  get 

Whole  area  =  p^  4-  pp*  |  — -  X~v~n  +  l 


—  n  •+•  r  ~  "  — 


1  —  n 


whence 

Area  of  Diagram  _  piVi(rl~*  —  w)  _  pl(rl~n  —  n~) 
rvl  "rt^CU^n)"          r(l  -  n)~ 

The  value  of  the  expansion  area  FCDE,  or  of  the  work  done  during 

expansion,  is  often  expressed  as^1-1  _  ^2  2.     This  value  is  imme- 

diately deducible  from  equation  (C),  thus: 

I"  (TV  V  ~  "  _  v  l  ~  "  ~i 
Expansion  area  =  p^i\n  i   _      l 


n  —  I  n  —  1  n  —  1 

—p.v, 
n  —  l 


9s  THE  ELEMENTS  OF  STEAM  ENGINEERING 

114.  Equation  (D)  is  the  general  expression  for  the  value  of  the 
mean  pressure,  whatever  the  value  of  n. 

In  reality,  n  is  the  ratio  of  the  specific  heats  of  the  gas  at  con- 
stant pressure  and  constant  volume,  and  has  a  yalue  between  unity 
and  1.405. 

If  steam  be  the  gas  under  consideration,  there  are  three  cases  : 

1.  Adiabatic  Expansion.  —  When  steam  expands  in   a  non-con- 
ducting and  non-radiating  cylinder,  so  that  it  neither  receives  nor 
parts  with  heat  during  the  expansion,  except  that  due  to  the  work 
performed,  the  expansion  curve  is  adiabatic  (without  transference). 
The  pressure  falls  as  a  result  of  the  increase  in  volume  and  in  the 
performance  of  the  external  work. 

The  law  governing  such  expansion  is  that  the  pressure  varies 
nearly  as  the  reciprocal  of  the  ninth  root  of  the  tenth  power  of  the 

volume;  or,  pec  -TO  ;  that  is,  pv}*  =  C  . 

This  case  is  useful  only  in  theoretical  investigations,  since  non- 
conducting and  non-radiating  cylinders  do  not  exist  in  practice. 

To  find  the  expression  for  the  mean  pressure  with  adiabatic  ex- 
pansion, let  n  =  ^°-  in  equation  (D),  and  we  shall  have: 

^(ri-V  -  W  _  Pi(r~*  -  V)      #(10  -  9r-*) 
Pm~      r(l  —  -1/)  _r  r 

9 

2.  Saturated  Steam  Expansion.  —  When  saturated  steam  expands 
in  a  cylinder,  doing  work,  but  receiving  from  a  jacket  or  other  ex- 
ternal source  sufficient  heat  to  prevent  liquefaction,  the  pressure  at 
all  points  of  the  stroke  is  that  due  to  the  volume  and  temperature  of 
saturated  steam. 

According  to  Eankine's  experiments  the  law  governing  the  expan- 

sion of  saturated  steam  is  that  p  <*  —  ^-  ;  that  is,  pv  &  =  C  . 
Letting  n  =  ^J  in  equation  (D),  we  have  : 

)  _  pl(r-A-ffi  _jPl(17-  16r-A) 


3.  Hyperbolic,  or  Isothermal,  Expansion.  —  When  damp  steam 
expands  in  a  cylinder,  doing  work  meanwhile  but  receiving  from 
some  external  source  an  amount  of  heat  equivalent  to  the  external 
work  performed,  the  expansion  curve  is  that  of  the  rectangular 


MEAN  PRESSURE  OF  GAS  EXPANDING  WITHIN  CYLINDER      99 

hyperbola,   and  the  pressure   varies   inversely   as   the   volume,   or 
P*\',  that  is,  pv  —  C. 

This  is  a  very  common  case  in  practice,  and  to  facilitate  calcula- 
tions, the  expansion  curve  of  the  indicator  diagram  is  assumed  to 
be  hyperbolic. 

If,  in  this  case,  we  substitute  directly  in  equation  (D)  the  value 
n  =  1,  the  result  is  $  ,  or  indeterminate;  but  if  we  evaluate  by 
differentiation  before  substituting  the  value  of  n,  we  get  the  correct 
result,  thus  : 

_  ffi(rl~n  —  ?0  _  d[pl(r't-H  —  n)']  __  j»i(  —  rl  ~  "  loge  rdn  —  dn  ) 
r(l  -  n)          ~d\r(\~—-ny\ 

1  +  r1—  log.r 

~  ~ 


Substituting  now  the  value  of  n,  we  have  : 

,  1  +  loger 
P-  =  Pi  X  -  -JJT*-  • 

Or  if,  in  the  expression: 

Whole  area  of  diagram  =  plvl  +  pj"  I     ^,  we  let  n  =  1,  we  have 

9/vl     v 

f*rv^dv  r       ~\rc\ 

Area  of  diagram  =  p^\  +  pp"  I       '—  =  plvl  -f  ppA  loge?; 

t/t'!  V  JUl 


=  ^^(1  -f-  loger),  whence 

__  Area  of  diagram  _plvl(\  +  loger)  _pi(l  +  log.,0 
rvj  rvj  r 

115.  To  construct  the  curves  representing  these  three  cases  of  ex- 
pansion, let  the  vertical  and  horizontal  lines  OA  and  OF,  Fig.  29, 
be  the  lines  of  pressures  and  volumes  respectively. 

Suppose  steam  of  200  pounds  absolute  pressure  per  square  inch 
to  be  admitted  into  a  cylinder  having  a  volume  of  10  cubic  feet,  and 
suppose  the  piston  to  be  driven  a  distance  AB,  equal  to  one-quarter 
stroke,  before  the  supply  of  steam  is  cut  off. 

There  will  be  -^  =  2.5  cubic  feet  of  steam  confined  in  the  cylin- 
der, which  will,  by  its  expansive  force,  drive  the  piston  to  the  end 
of  the  stroke. 

We  will  suppose  the  conditions  of  the  expansion  to  be  hyperbolic, 
so  that  the  pressures  will  be  inversely  as  the  volumes,  according  to 


100 


THE  ELEMENTS  OF  STEAM  ENGINEERING 


Boyle's  law.     When  the  steam  is  expanded  to  5  cubic  feet,  or  twice 
its  initial  volume,  the  pressure  will  have  fallen  to  one-half  the  initial 

pressure  =  -~-  =  100  Ibs.  =  MN.    When  the  expansion  has  been 

carried  to  7.5  cubic  feet,  or  to  three  times  the  initial  volume,  the 

20 
pressure  will  be  reduced  to  one-third  the  initial  pressure  =  -=- 

—  66|  Ibs.  =  GK.    When,  finally,  the  piston  reaches  the  end  of  the 


cubic  f&ei 

FIG.  2g. 

stroke,  the  volume  of  steam  will  be  10  cubic  feet,  or  four  times  the 
initial  volume,   and  the  pressure,  therefore,   will  have   fallen  to 

900 

~    =  50  Ibs.  =  FC.     If  the  points  B,  N,  K,  and  C  be  joined,  the 
curve  will  be  that  of  the  equilateral,  or  rectangular,  hyperbola. 

The  rectangles  OB,  ON,  OK,  and  OC,  represent  the  conditions 
of  the  steam  as  to  volume  and  pressure  when  the  piston  is  at  the 
positions  JB,  MN,  GK,  and  FC  respectively,  and  these  rectangles 


MEAN  PRESSURE  OF  GAS  EXPANDING  WITHIN  CYLINDER     101 

are  all  equal  in  area;  for  200  X  2.5  =  100  X  5  =  66f  X  7.5 
=  50  X  10  =  500. 

A  sufficient  number  of  points  were  not  obtained  to  draw  the  curve 
with  any  degree  of  accuracy,  but  we  may  easily  obtain  as  many 
points  as  we  please. 

116.  The  rectangular  hyperbola  may  easily  be  described  with- 
out any  calculation,  thus: 

Complete  the  parallelogram,  OW,  Fig.  29.  Take  any  point,  as 
T,  in  the  line  of  volumes,  and  draw  T8  perpendicular  to  OF.  Join 
SO  cutting  JB,  the  vertical  at  the  point  of  cut-off,  at  R.  Through 
R  draw  HR  parallel  to  the  line  of  volumes  OF,  and  produce  it  until 
it  intersects  the  vertical  TS  at  P;  then  P  is  a  point  of  the  curve. 

117.  To  plot  the  Saturation  Curve  pv&  =  C.  we  may  proceed  as 
follows:  p  —  -g,  consequently  log  p  —  log  C '  —    —•*£-  =  l°g  G 

— 1.0625  log  v.  If  vv  the  volume  at  cut-off,  be  denoted  by  1,  we  shall 
have  C  =  p^v^  =  p± .  Having  now  a  value  of  (7,  we  shall  have  by 
substitution  log  p  =  log  p^  —  1.0625  log  v,  by  means  of  which  we 
obtain  the  saturation  curve  BD  of  Fig.  29,  as  follows : 

The  initial  volume  A  B,  or  volume  up  to  cut-off,  being  denoted  by 

/» 

unity,  the  volume  OQ  must  be  denoted  by  ~-F  =  2.4,  therefore, 

log  p  =  log  200  —  1.0625  log  2.4. 

2.4  log  0.38021  llog  9.58002  —  10 
1.0625  log  0.02633 

(1.0625  log  2.4)  log  0.40397  llog  9.60635  —  10 
200  log  2.30103 

p  log  1.89706 
Therefore,  p  =  78.9  =  QL '. 

In  like  manner  the  Adiabatic  Curve,  pv1*  =  (7,  may  be  plotted. 

TOT   i,  C      ,  ,      n      10 log?; 

We  have  p  —  -^>,  whence  log  p  =  log  C  —  — <p- 

=  log  C  —  1.111  logv. 

If  we  let  v±  =  1,  we  shall  have  log  p  =  log  p±  —  1.111  log  v,  by 
means  of  which  we  obtain  the  adiabatic  curve  BE  of  Fig.  29,  as 
follows : 


102  THE  ELEMENTS  OF  STEAM  ENGINEERING 

log  p  —  log  200  —  1.111  log  2.4 

2.4  log  0.38021  Hog  9.58002  —  10 
1.111  log  0.04571 

(1.111  log  2.4)  log  0.42240  Hog  9.62573  —  10 
200  log  2.30103 

p  log  1.87863 
Therefore,  p  =  75.62  =  QL" . 

Figure  29  affords  a  ready  means  of  comparing  the  three  expansion 
curves.  They  represent  the  expansions  under  the  respective  con- 
ditions of  an  initial  volume  AB  of  steam  of  absolute  pressure 
OA  =  p19  to  a  final  volume  OF,  and  a  final,  or  terminal,  pressure 
p2  -=2  FC,  FD,  and  FE,  respectively.  The  three  curves  will  always 
be  relatively  as  shown. 

PEOBLEMS. 

33.  Absolute  initial  pressure,  88  Ibs.  per  sq.  inch;    ratio  of  ex- 
pansion, 4.     Find  the  terminal  pressures  for  the  three  cases  of  ex- 
pansion, viz. :   Adiabatic,  Saturation,  and  Hyperbolic. 

Ans.  18.660  Ibs.;   20.174  Ibs.;   22.000  Ibs. 

34.  Find  the  mean  and  terminal  pressures  of  steam  at  160  Ibs. 
absolute  pressure,  after  expanding  4  times,  and  when  the  conditions 
of  the  expansion  are:   (a)   Adiabatic;  (&)   Saturated;   (c)  Hyper- 
bolic. Ans.   (a)  91.40  and  34.33;  (6)  93.12  and  36.70; 

(c)  95.44  and  40.00. 


CHAPTER  X. 
BOILER  AND  ENGINE  EFFICIENCY. 

118.  Boiler  Efficiency. — The  question  of  boiler  and  engine  effi- 
ciency has  already  been  adverted  to  in  Chapter  IV,  and  it  is 
here  the  purpose  to  consider  it  more  fully. 

The  efficiency  of  a  boiler  is  usually  expressed  in  terms  of  the 
weight  of  steam  it  produces  per  pound  of  coal  it  consumes,  though 
the  efficiency  in  certain  circumstances  may  very  properly  be  ex- 
pressed in  terms  of  the  steam  production  with  respect  to  the 
weight'  or  to  the  first  cost  of  the  boiler. 

In  estimating  the  efficiency  of  a  boiler  under  the  basis  of  coal 
consumption,  all  heat  due  to  the  coal  thrown  into  the  furnace  is 
charged  to  the  boiler,  while  it  is  credited  only  with  the  heat  trans- 
ferred through  the  heating  surface  and  used  in  the  generation  of 
steam.  The  ratio  of  these  two  quantities  expresses  the  efficiency 
of  the  boiler.  If  the  two  quantities  were  equal  the  efficiency  would 
be  perfect  and  be  expressed  by  unity,  but  owing  to  various  causes 
the  actual  efficiency  varies  from  0.55  to  0.75. 

There  are  a  number  of  sources  to  which  this  loss  in  efficiency 
may  be  attributed.  In  the  first  place  the  losses  in  the  furnace 
under  the  most  favorable  conditions  vary  from  2.5  to  4  per  cent. 
The  coal  may  be  such  that  portions  of  it  fall  through  the  grate 
unconsumed,  and  oftentimes  the  finer  portions,  when  only  partly 
consumed,  are  carried  from  the  furnace  by  the  draft  and  lodged 
in  the  tubes  and  uptakes.  The  supply  of  air  at  times  may  be  in- 
sufficient for  complete  combustion,  resulting,  as  was  stated  on 
page  24,  in  the  production  of  carbonic  oxide  (CO)  of  a  thermal 
value  of  only  4500  units  per  pound,  or  less  than  one-third  that  due 
perfect  combustion.  Imperfect  combustion  produces  smoke  con- 
taining carbon  only  partly  consumed  and  is,  of  course,  lost  through 
the  smoke  pipe. 

All  the  heat  which  is  liberated  in  the  furnace  gases  is  not,  by 
any  means,  transmitted  through  the  heating  surface  to  the  water 
in  the  boiler.  A  small  part  is  radiated  to  the  surrounding  atmos- 
phere, and  another  and  much  larger  part  escapes  through  the 
smoke  pipe.  The  heat  contained  in  the  gases  escaping  through  the 


104  THE  ELEMENTS  OF  STEAM  ENGINEERING 

smoke  pipe  is  the  most  serious  loss  of  all  and  is  inevitable,  as  its 
prevention  could  only  be  accomplished  by  reducing  the  temperature 
of  the  chimney  gases  to  that  of  the  outside  air.  This,  of  course, 
cannot  be  done,  since  the  temperature  of  the  gases  must  be  higher 
than  that  of  the  water  in  the  boiler  to  prevent  a  transference  of 
heat  in  the  wrong  direction.  The  temperature  of  the  escaping 
chimney  gases  will  always  be  considerably  higher  .than  that  of  the 
water  in  the  boiler,  for  in  no  instance  would  sufficient  heating  sur- 
face be  given  to  a  boiler  to  reduce  their  temperature  to  that  of  the 
steam.  Aside  from  this,  the  natural  draft  of  the  boiler,  occasioned 
by  the  difference  of  weights  between  the  columns  of  air  within  and 
without  the  smoke  pipe,  depends  directly  upon  the  temperature  of 
the  chimney  gases,  and  this  practical  consideration  fixes  the  tem- 
perature between  the  limits  of  500°  and  600°. 

Under  the  most  favorable  conditions  the  losses  above  considered 
aggregate  little  less  than  25  per  cent,  and  under  the  conditions  of 
insufficient  air  for  combustion  and  incompetent  firing  the  aggregate 
may  reach  as  much  as  45  per  cent. 

In  estimating  the  evaporative  efficiency  of  boilers  the  condition 
of  the  steam  with  respect  to  its  dryness  should  be  ascertained  in 
order  to  avoid  erroneous  results.  In  practice,  the  steam  produced 
by  a  boiler  contains  more  or  less  moisture  held  in  suspension,  and 
it  is  evident  that  the  total  heat  required  to  produce  such  steam  is 
less  than  would  be  required  to  produce  saturated  steam,  to  the 
extent  of  the  latent  heat  which  would  be  required  to  convert  the 
suspended  moisture  into  steam.  For  example,  suppose  a  boiler  to 
produce  dry  steam  at  a  temperature  of  366°  from  a  feed  water  tem- 
perature of  122°.  Then  the  total  heat  to  produce  one  pound  of 
the  steam  would  be  Hw  =  1091.7  +  0.305(366  —  32)— (122  —  32) 
=  1103.57  units. 

If  the  steam  produced  by  the  boiler  contained  5  per  cent  of  sus- 
pended moisture,  instead  of  being  dry,  then  the  heat  required  to 
produce  a  pound  of  the  moist  steam  would  be :  Hw  =0.95HL 
+  H8  —  Hf  =  0.95[1091.7  —  0.7(366  --  32)]  +  (366  —  32) 
—  (122  —  32)  =  1059  units. 

If,  in  this  instance,  the  5  per  cent  of  moisture  were  neglected 
and  the  steam  assumed  to  be  dry,  the  evaporative  efficiency  would 

be  exaggerated  in  the  ratio  1Q^9  =  1.042,  or  to  the  extent  of 
4.2  per  cent. 


BOILED  AND  ENGINE  EFFICIENCY  105 

In  the  above  example  the  sensible  heat,  Hs,  or  the  heat  in  the 
steam  at  the  boiling  point,  is  taken  as  t  —  32,  in  which  t  is  the 
temperature  of  the  steam  at  the  pressure  considered.  This  value 
is  less  than  those  given  in  the  table  of  the  properties  of  saturated 
steam,  and  the  difference  arises  from  the  fact  that  in  the  value, 
t  —  32,  the  assumption  is  made  that  one  degree  rise  in  temperature 
per  pound  of  water  is  the  equivalent  of  one  thermal  unit.  This  is 
not  exactly  true,  for  at  temperatures  above  the  standard  at  which 
the  B.  T.  U.  was  measured  (39°,  or  more  recently  62°)  the  water 
absorbs  a  small  quantity  of  heat  due  to  the  work  performed  in 
slightly  expanding  it.  The  sensible  heats  given  in  the  table  include 
this  small  quantity,  and  are,  therefore,  equal  to  t  —  32  plus  the 
heat  converted  into  work  in  expanding  the  water.  The  difference 
is  not  great,  however,  and  in  the  absence  of  tables  the  sensible  heats 
may  be  taken  as  t  —  32.  For  the  same  reason  the  table  values  of 
the  latent  heat,  or  the  heat  of  vaporization,  of  steam  are  slightly 
smaller  than  those  derived  from  the  formula  H£  =  1091.7 

—  0.7  (£  —  32),  since  the  table  values  are  derived  by  subtracting  the 
sensible  heat  from  the  total  heat  (HL  =  HT  —  Hs) . 

In  the  examples  which  follow,  the  table  values  for  the  latent  and 
sensible  heats  will  be  taken,  though  in  the  absence  of  tables  they 
may  be  calculated  without  any  material  error  in  results. 

119.  We  may  modify  the  formula  for  finding  the  total  heat  re- 
quired for  the  evaporation  of  one  pound  of  water  (Hw)  so  as  to 
meet  the  cases  where  the  condition  of  the  steam  with  respect  to  its 
dryness  is  to  be  taken  into  account.     Thus,  if  x  denotes  the  "dry- 
ness  -fraction"  of  the  steam  we  will  then  have  Hw  =  xHL-\-H8  — Hf. 
If  the  steam  is  saturated,  or  dry,  x=I9  and  the  formula  becomes 
Hw  —  HL  +  H8—Hf  =  H,—  Hf  —  1091.7  +  0.305(£  —  32) 

—  (tf  —  32),  which  is  the  formula  previously  given  for  Hw. 

120.  The  dryness  fraction  of  steam  may  be  determined  by  a 
calorimeter  test.     There  are  three  forms  of  calorimeters  in  use  for 
this  purpose,  known  as  the  Barrel,  the  Separating,  and  the  Throt- 
tling calorimeters.     The  barrel  calorimeter  is  the  simplest,  and  if 
great  care  is  exercised  in  the  tests  the  results  obtained  are  nearly  as 
accurate  as  those  derived  from  the  other  and  more  complicated 
contrivances. 


106  THE  ELEMENTS  OF  STEAM  ENGINEERING 

121.  The  Barrel  Calorimeter  consists  of  a  barrel  of  hard  wood, 
about  three-fourths  filled  with  water  and  placed  on  accurate  plat- 
form scales.  A  half -inch  pipe  leads  from  the  main  steam  pipe  of 
the  boiler,  the  part  of  the  pipe  within  the  steam  pipe  being  closed 
at  the  end  and  perforated  with  holes  about  one-eighth  inch  in  diam- 
eter. This  pipe  should  be  covered  with  some  non-conducting  sub- 
stance, and  should  be  thoroughly  heated  before  commencing  the 
test.  The  steam  is  admitted  to  the  barrel  through  a  hose  attached 
to  the  half-inch  pipe.  This  hose  reaches  nearly  to  the  bottom  of 
the  barrel,  is  closed  at  the  end  and  perforated  for  about  9  inches 
of  its  length  from  the  end.  During  the  entrance  of  the  steam  the 
hose  should  be  moved  about  so  as  to  promote  the  mixture  of  the 
condensed  steam  and  the  condensing  water.  When  the  scales  show 
that  the  desired  weight  of  steam  (about  one-fifteenth  of  the  original 
weight  of  the  water  in  the  barrel)  has  entered  the  barrel,  the  re- 
sulting temperature  of  the  water  is  taken  by  a  thermometer  deeply 
immersed.  The  pressure  of  the  steam  is  taken  from  the  gauge, 
and  the  corresponding  temperature  and  latent  heat  is  taken  from 
the  table.  The  temperature  of  the  water  in  the  barrel,  before  and 
after  the  mixture,  should  be  taken  with  great  care  and  with  a  ther- 
mometer graduated  to  at  least  one-quarter  of  a  degree,  and  it  is 
not  advisable  to  have  the  temperature  of  the  mixture  much  over 
110°  in  order  that  radiation  may  be  avoided. 

Let  W  =  original  weight  of  water  in  barrel, 

W^  =  weight  of  steam  and  suspended  moisture  blown  in, 
w  =  weight  of  dry  steam  supplied, 
t  =  original  temperature  of  water  in  barrel, 
tj_  =  temperature  of  steam, 

t2  =  temperature  of  the  water  in  barrel  after  the  addition 
of  the  steam. 

The  heat  lost  by  the  steam  will  be  the  latent  heat  of  the  dry 
steam  supplied  =  wHL ,  plus  the  heat  given  up  in  cooling  the  lique- 
fied dry  steam  and  the  water  (moisture)  carried  in  by  the  steam 
from  the  temperature  ^  to  the  temperature  t2.  This  latter  quan- 
tity is  equal  to  W^(t±  —  t2). 

The  total  heat  lost  by  the  steam  will  then  be : 
wHL+Wi(t,  —  t2). 

The  heat  gained  by  the  water  will  be  W(t2  —  t). 


BOILER  AND  ENGINE  EFFICIENCY  107 

If  the  heat  due  to  cooling  the  liquefied  steam  and  the  water 
(moisture)  carried  in  it,  be  subtracted  from  the  total  heat  lost, 
the  remainder  will  be  the  heat  due  to  the  latent  heat  of  the  dry 
steam  supplied,  and  we  shall  have  wHL  —  W(t2  —  t)  —  W^^  —  t2), 


—  i  .  ,       .  , 

whence  w  =  —  —  -  —jf  —  s  -  =  weight  of  dry  steam  sup- 
**i 

plied.     Calling  y  the  fraction  of  moisture  in  the  steam,  we  shall 

(Wt-iti 

have  y  —  -  —  ™-,  —  • 

And  if  x  denotes  the  dryness  fraction  of  the  steam,  we  shall  have 

w 
x==l-y=Wi.      , 

EXAMPLE  I.  —  The  barrel  of  a  calorimeter  contains  175  pounds  of 
water  at  a  temperature  of  60°.  After  8  pounds  of  moist  steam  at 
an  absolute  pressure  of  165  pounds  had  been  blown  into  the  barrel 
the  temperature  of  the  mixture,  was  found  to  be  108°.  Find  the 
dryness  fraction  of  the  steam. 

Solution.  —  From  the  tables  the  temperature  of   steam  at   165 

pounds  pressure  is  found  to  be  366°,  and  the  latent  heat  855.6  units. 

W(t,  —  /)  —  W1(tl  -  t,)  _  175(108  —  60)  -  8(366  -  108) 

H~T  855.6 

=  7.405  pounds  of  dry  steam  supplied. 

Then,  dryness  fraction  =  —  =  ?4^  =  0.9256 

FT  1  O 

122,  Throttling  Calorimeter.  —  The  operation  of  this  calorimeter 
depends  upon  the  principle  that  the  increase  in  volume  of  steam, 
unaccompanied  by  the  performance  of  work,  occasions  a  liberation 
of  heat  which  may  be  utilized  in  the  evaporation  of  any  moisture  the 
steam  may  contain  and  in  raising  the  temperature  of  the  steam 
above  that  due  to  its  pressure;  that  is,  superheat  it. 

The  form  of  this  calorimeter  designed  by  Prof.  E.  C.  Carpenter, 
and  manufactured  by  the  Shaeffer  and  Budenberg  Co.,  is  shown  in 
Fig.  30.  It  consists  of  a  collecting  nipple  screwed  into  the  main 
steam  pipe;  the  calorimeter  proper,  a  vessel  to  which  steam  is  ad- 
mitted through  a  converging  orifice,  and  containing  a  deep  cup  filled 
with  oil,  into  which  a  thermometer  is  inserted  to  obtain  the  tem- 
perature of  the  steam  within  the  calorimeter;  a  manometer 
attached  to  the  calorimeter  for  determining  the  pressure  within 
the  interior  ;  a  globe  valve  in  the  pipe  admitting  steam  to  the  calori- 
meter, used  merely  to  throttle  the  steam  and  thus  regulate  its  pres- 


108 


THE  ELEMENTS  OF  STEAM  ENGINEERING 


sure  in  the  calorimeter ;  and  a  pipe  to  carry  off  the  steam  from  the 
calorimeter. 

Let  PI  be  the  boiler  pressure  taken  from  a  gauge  near  the  calori- 
meter, HL  the  latent  heat  of  vaporization,  Hs  the  sensible  heat 
corresponding  to  p±,  and  x  the  weight  of  dry  steam  in  one  pound 
of  the  mixture  as  taken  from  the  boiler.  Then  1  —  x  denotes  the 
weight  of  moisture  in  a  pound  of  the  mixture,  and  xHL  -\-Hg  is 
the  total  heat  in  one  pound  of  the  mixture. 

Let  pz  be  the  pressure  in  the  calorimeter,  and  H*  and  tz  the  total 
heat  and  temperature  of  saturated  steam  corresponding  to  p2.  Let 


THERMOMETER 


Fig.  30. 


t  be  the  temperature  of  the  superheated  steam  in  the  calorimeter, 
as  indicated  by  the  thermometer.  Then,  the  heat  in  one  pound  of 
the  steam  in  the  calorimeter  is  : 


in  which  the  constant  0.48  is  the  specific  heat  of  superheated  steam. 
Assuming  that  there  is  no  heat  lost  from  radiation,  we  shall  have  : 
xHL  +  Ha  =  H?  +  0.48(7  —  tt),  whence 
Hf  +  0.48(1  —  tj  -  If, 

X    —    -  TT  > 

«* 

and   the   percentage    of   moisture   in    the   boiler    steam   will    be 


The  values  for  Hu  H8,  H*  and  iz  are  to  be  taken  from  the  table 
of  the  properties  of  saturated  steam. 


OF  THE 

UNIVERSITY 

BOILER  AND  ENGINE  EFFICIENCY 


EXAMPLE. — The  following  data  were  taken  from  a  test : 
Absolute  steam  pressure,  155  pounds.    Absolute  pressure  in  calori- 
meter, 14.8  pounds.     Temperature  of  steam  in  calorimeter,  260.4°. 
Prom  this  we  obtain : 

-  H46.7  +  0.48(260.8  —  212)  -.  332.7 
859.3 

=  0.975  Ib.  of  dry  steam  in  one  pound  of  the  boiler  steam. 

Percentage  of  moisture  in  steam  =  100(1- — 0.975)  =2.5  per 
cent. 

If  is  obvious  that  if  the  heat  liberated  fails  to  superheat  the  steam 
there  can  be  no  determination  of  its  quality  with  the  throttling 
calorimeter.  The  limit  beyond  which  this  form  of  calorimeter  is 
inoperative  is  when  t  and  t2  are  equal ;  that  is,  when  the  heat  liber- 
ated is  just  sufficient  to  evaporate  the  moisture  in  the  steam,  leaving 
it  dry  and  saturated  but  with  no  degree  of  superheat.  The  practical 
limit  is  when  the  steam  contains  little  in  excess  of  2.5  per  cent  of 
moisture,  and  the  most  convenient  pressure  to  be  maintained  within 
the  calorimeter  is  that  when  the  manometer  registers  at  the  zero 
mark,  or  when  the  absolute  pressure  is  that  of  the  atmosphere,  which 
may  be  taken  as  14.7  pounds  in  the  absence  of  a  barometer.  If  the 
reading  of  the  barometer  is  taken,  the  pressure  is  obtained  by  multi- 
plying by  0.49. 

123.  Separating  Calorimeter. — For  the  determination  of  the 
quality  of  steam  containing  a  degree  of  moisture  beyond  the  limit 
of  the  throttling  calorimeter,  some  form  of  steam  separator  is  used. 
The  form  largely  in  use  is  that  devised  by  Prof.  Carpenter.  This 
instrument  provides  for  a  mechanical  separation  of  the  moisture 
from  the  steam,  which  permits  the  two  quantities  to  be  weighed 
separately. 

124  In  making  tests  of  boilers,  allowance  is  made  for  the  moisture 
and  ash  in  the  coal,  in  order  that  the  evaporative  power  of  a  pound 
of  the  combustible  may  be  obtained. 

EXAMPLE. — A  boiler  produces  steam  at  175  pounds  pressure  per 
gauge  from  a  feed  water  temperature  of  142°,  evaporating  8.5  Ibs. 
of  water  per  pound  of  coal.  The  dryness  fraction  of  the  steam  is 
0.98.  The  coal  contains  10  per  cent  of  ash  and  5  per  cent  of  mois- 
ture. Find  the  actual  evaporation,  and  the  equivalent  evaporation 
from  and  at  212°,  of  a  pound  of  the  dry  combustible. 


110  THE  ELEMENTS  or  STEAM  ENGINEERING 

Here  we  have : 

Hw  a=  xHL+  Hs  —  Hf  =  0.98  X  847  +  350  --  (142  --  32) 
=  1070  units.  Equivalent  evaporation  from  and  at  212°  is 

8.5  X  1070       Q/I01, 
-m—  =  9.42  Ibs. 

The  evaporations  of  8.5  pounds  and  9.42  pounds  were  obtained 
from  0.85  pound  of  dry  combustible,  therefore  the  actual  and  equiva- 

o   o 

lent  evaporations  of  a  pound  of  dry  combustible  are  Q-^T  =  10  Ibs., 

9  42 
and  ^-— •  =i  11.08  Ibs.,  respectively. 

125.  Engine  Efficiency. — Mention  has  already  been  made  of  Car- 
not' s  Eeversible  Cycle,  or  Perfect  Heat  Engine.  In  practice  the 
steam  engine  transforms  only  a  portion  of  the  mechanical  energy 
of  the  heat  contained  in  the  steam  supplied  to  it.  The  greater  the 
portion  of  the  heat  transformed  into  work  the  greater  the  efficiency 
of  the  engine,  and  it  is  the  constant  aim  of  the  engineer  to  make 
this  portion  as  great  as  possible. 

Unfortunately  there  are  a  number  of  causes  which  reduce  the 
heat  converted  into  useful  work  to  a  small  fraction  of  the  heat  sup- 
plied. The  principle  of  these  causes  is  the  Second  Law  of  Thermo- 
dynamics, which  asserts  that,  "  Heat  cannot  pass  from  a  cold  body 
to  a  hot  one  by  a  purely  self-acting  process."  It  follows  from  this 
law  that  no  steam  engine  can  convert  into  work  all  the  heat  of  the 
steam  supplied  to  it,  for  as  soon  as  the  temperature  of  the  steam 
falls  to  that  of  the  surrounding  atmosphere,  the  heat  remaining  in 
it  is  not  available  for  doing  useful  work.  Thus,  if  T^  be  the  highest 
absolute  temperature  available,  and  T2  the  lowest,  then  the  maxi- 

T  —  T 
mum  efficiency  obtainable  is   "^m — ? ,  or  if  tf±  be  the  temperature  of 

the  steam  of  initial  pressure,  and  t2  the  temperature  of  the  steam  of 
final  pressure,  then  the  maximum  efficiency  of  the  most  perfect 
engine  is 

ft  +  461)  -  (I,  +  461)  _  t,  -  tt 
T,  -~7T 

As  an  illustration,  suppose  an  engine  working  between  the  tem- 
peratures of  366°  and  142°,  corresponding  to  initial  and  final  pres- 
sures of  165  Ibs.  and  3  Ibs.  absolute,  respectively.  The  ideal  effi- 

q  ft  ft  -«    A  f\ 

ciency  would  then  be  nt^rrzar  =  0.27,  or  little  more  than  25  per 
cent. 


BOILER  AND  ENGINE  EB^EICIENCY  111 

126.  The  cycle  of  operations  of  this  ideal  engine,  known  as  the 
Carnot  cycle,,  consists  of  four  stages,  specified  with  simplicity  by 
Professor  W.  F.  Durand,  in  his  "  Practical  Marine  Engineering/'  as 
follows  : 

(1)  The  first  operation  must  consist  of  an  expansion  at  constant 
temperature,  and  all  heat  received  from  the  source  of  supply  must 
be  received  during  this  operation  (Isothermal  expansion). 

(2)  The  second  operation  must  consist  of  an  expansion  with  de- 
crease of  temperature,  during  which,  however,  no  heat  is  allowed  to 
enter  or  leave  the  substance  (Adiabatic  expansion). 

(3)  The  third  operation  must  consist  of  a  compression,  during 
which  the  temperature  remains  constant,  and  all  heat  removed  from 
the  body  must  be  removed  during  this  operation  (Isothermal  com- 
pression) . 

(4)  The  fourth  operation  must  consist  of  a  compression  with  in- 
crease of  temperature,  during  which,  however,  no  heat  is  allowed  to 
enter  or  leave  the  substance,  and  at  the  end  the  substance  must  find 
itself  in  the  same  condition  as  at  the  beginning  of  number   (1) 
(Adiabatic  compression). 

It  will  be  observed  that  work  is  done  by  the  substance  during 
operations  (1)  and  (2),  and  that  work  is  done  on  the  substance  dur- 
ing (3)  and  (4).  The  difference  between  work  done  by  and  on  the 
substance  will  be  the  net  work  obtained  from  the  heat  in  the  sub- 
stance, and  the  ratio  of  this  to  the  total  heat  supplied  during  (1) 
is  the  efficiency  which  will  be  exactly  measured  by  the  difference  of 
the  absolute  temperatures  of  operations  (1)  and  (3),  divided  by 
the  absolute  temperature  of  (  1  )  ;  that  is  : 

Efficiency  of  the  Ideal  Engine  = 


127.  The  conditions  of  the  four  operations  of  the  ideal  engine  are 
far  from  being  obtained  in  practice,  and  under  the  most  favorable 
circumstances  not  more  than  from  60  per  cent  to  70  per  cent  of  the 
ideal  efficiency  is  obtained  in  actual  practice.  The  principal  causes 
which  reduce  the  efficiency  of  the  steam  engine  below  that  of  the 
ideal  engine  are  enumerated  on  page  34. 

Within  the  present  limits  of  temperature  used  in  the  steam  engine 
the  ideal  efficiency  varies  from  25  to  30  per  cent,  while  the  actual 
efficiency  varies  from  15  to  20  per  cent. 


THE  ELEMENTS  OF  STEAM  ENGINEERING 


128.  Weight  of  Steam  per  I.  H.  P.  per  Hour.  —  In  the  ideal  engine 
the  efficiency  is    v--  -T^TO,  and  since  one  I.  H.  P.  requires  an  ex- 

?1    ~T~    401 

penditure  of      *  g  -  =  42.42   thermal  units,   it  follows  that  the 
minimum  expenditure  in  the  ideal  engine  to  produce  one  I.  H.  P. 


will  be  42,42  +-  = 


46  ^ 


thermal  units  per  minute. 


If  Hw  denotes  the  units  of  heat  required  to  produce  one  pound 
of  steam  from  the  temperature  t.2  and  at  the  temperature  tlt  and  N 
denotes  the  number  of  pounds  of  steam  required  per  I.  H.  P.  per 
hour,  we  will  have  : 


„_  60[42.42(^  +  461)]  _ 
^~~~~     ~ 


60[42.42«1  +  461)] 


_  _  __ 

[1091.7  +  0.305^  —  32)  —  «3  —  32)]  [^  —  <a]  ' 

The  consumption  of  steam  per  hour  of  the  ideal  engine,  both 
condensing  and  non-condensing,  given  in  the  following  table,  were 
calculated  by  the  above  formula.  In  the  case  of  the  condensing 
engine  the  feed  water  is  assumed  to  have  been  taken  from  the  con- 
denser at  a  temperature  of  100°,  and  in  the  non-condensing  engine 
it  is  assumed  that  the  exhaust  steam  has  been  used  to  raise  the  tem- 
perature of  the  feed  to  about  212°. 

The  table  shows  a  marked  gain  of  efficiency  due  to  the  use  of  the 
condenser,  and  this  gain  is  greater  for  low  than  for  high  pressures. 

POUNDS  OF  STEAM  PER  I.  H.  P.  PER  HOUR. 


Initial  pressure 
in  pounds. 

Condensing 
Engine. 

Non-condensing 
Engine. 

tj  Fahr. 

70 

8.62   Ibs. 

19.32  Ibs. 

303° 

90 

8.13   Ibs. 

16.55  Ibs. 

320° 

110 

7.73  Ibs. 

14.76  Ibs. 

335° 

1'30 

7.43  Ibs. 

347° 

150 

7.19   Ibs. 

358° 

180 

6.89  Ibs. 

373° 

Since  the  actual  efficiencies  of  steam  engines  vary  from  60  to  70 
per  cent  of  the  ideal  efficiencies,  it  will  be  seen  from  the  table  that 
a  consumption  of  from  11.5  to  12  Ibs.  of  steam  per  I.  H.  P.  per 


BOILER  AND  ENGINE  EFFICIENCY  113 

hour  may  be  expected  at  pressures  commonly  used  with  triple-ex- 
pansion engines,  and  in  the  case  of  non-condensing  engines  a  con- 
sumption of  25  pounds  per  I.  H.  P.  per  hour  may  be  expected  under 
favorable  conditions.  Assuming  the  boiler  to  evaporate  9.5  pounds 
of  water  per  pound  of  coal,  the  above  figures  would  indicate  a  coal 
consumption  of  1.2  pounds  per  I.  H.  P.  per  hour  in  the  case  of  the 
triple-expansion  engines  and  2.7  pounds  in  the  case  of  the  non- 
condensing  engines.  These  figures  conform  fairly  well  with  the 
results  obtained  in  actual  practice. 

129.  Liquefaction  in  Cylinders— On  page  37  the  object  of  jack- 
eting cylinders  was  pointed  out.  It  is  not  probable  that  the  heat 
from  the  j-acket  steam  ever  entirely  prevents  initial  condensation, 
or  supplies  heat  sufficient  to  replace  all  that  transmuted  into  work 
during  expansion,  and  thus  maintain  the  steam  in  a  state  of  satura- 
tion, but  there  can  be  no  question  as  to  the  advantage  derived  from 
jacketing  the  cylinders  of  engines  of  slow  and  moderately  high 
speeds. 

Experiments  have  shown  that  the  use  of  the  jacket  in  such  in- 
stances increased  the  efficiency. of  the  steam  from  6  to  25  per  cent. 

It  is  true  that  the  transference  of  heat  from  the  jacket  to  the 
cylinder  is  accompanied  by  a  corresponding  liquefaction  in  the 
jacket,  and  it  would  appear  that  the  scene  only  of  the  liquefaction 
has  been  changed,  but  the  essential  difference  is  that  the  steam 
liquefied  in  the  jacket  passes  off  in  the  state  of  water  at  the  tem- 
perature of  the  jacket  steam,  each  pound  carrying  with  it  a  number 
of  units  equal  to  the  sensible  heat  of  the  steam,  while  that  liquefied 
in  the  cylinder  escapes  to  the  air  or  condenser  in  the  state  of  steam, 
carrying  with  it  the  latent  heat  of  the  jacket  steam.  The  water 
from  the  jacket  is  invariably  drawn  off  and  returned  to  the  boiler 
with  the  feed,  and  the  heat  supplied  to  the  cylinder  per  pound  lique- 
fied in  the  jacket  is  the  latent  heat  of  the  steam  corresponding  to  the 
pressure  of  the  jacket  steam. 

The  following  example  presents  the  theoretical  consideration  of 
the  action  of  steam  in  a  jacketed  cylinder. 

A  simple  expansive  engine  has  a  jacketed  cylinder  9"  in  diameter, 
and  a  stroke  of  10".  Cut-off,  0.25  stroke;  clearance,  6  per  cent; 
initial  pressure  of  steam,  95  Ibs.  absolute;  back  pressure,  18  Ibs.; 
revolutions  per  minute,  375.  The  specific  volume  of  steam  at  95  Ibs. 
pressure  is  4.62  cubic  feet,  and  it  is  assumed  that  only  0,9  of  the 
8 


114  THE  ELEMENTS  OF  STEAM  ENGINEERING 

theoretical  mean  pressure  is  realized.  Find:  (1)  The  weight  of 
working  steam  per  I.  H.  P,  per  hour;  (2)  the  weight  of  jacket 
steam  per  I.  H.  P.  per  hour,  and  its  equivalent  from  the  tempera- 
ture of  feed  water;  (3)  the  efficiency  of  the  engine;  (4)  the  total 
weight  of  steam  per  I.  H.  P.  per  hour. 

Solution.—  Stroke  =  |f  =  $  feet.     10  X  0.06  =  0.6  inch  equiva- 
lence of  clearance  in  stroke.     One-fourth  stroke  =  2.5  inches,  and 

'  —  i—  :-  =  -nr  feet  of  stroke  up  to  cut-off,  including  clearance. 

9 
Eadius  =  0  .  .  1tl  =  f  feet. 

<i  X  1/& 
3>1416  *^3JX—  =  0.368  cu.  ft.  volume  swept  by  piston  per  stroke. 

3'(8)'6>Tl28X  *62  "  =  °-02473  lb'  steam  used  Per  stroke' 
Eatio  of  expansion  =     25++°b°06  =  3'42  loge 


=61921b 

6Aa 

Theoretical  m.  e.  p.  =  61.92  —  18  =  43.92  Ibs. 
Mean  effective  pressure  to  be  expected  =  43.92  X  0.9  =  39.528  Ibs. 
Work  per  stroke  =  Pv  —  144  X  39.528  X  0.368  =  2094.7  foot  Ibs. 

90Q4-  7 
Work  per  pound  of  steam  =002473=  84>704  foot  lbs> 


qq 

Steam  per  I.  H.  P.  per  hour  =       g4  7  =  23'37  lbs* 

Or,  the  weight  of  steam  per  I.  H.  P.  may  be  found  thus  : 
Weight  of  steam  used  per  hour  =  0.02473  X  750  X  60  =  1112.85 
Ibs. 

2094.7X750         .„  „ 


Steam  per  I.  H.  P.  per  hour  =          '      =  23.37  Ibs. 

The  steam  in  the  jacket  supplies  the  heat  necessary  to  maintain 
the  working  steam  in  a  state  of  saturation  throughout  the  stroke, 
and  the  measure  of  this  heat  is  the  latent  heat  of  the  steam  liquefied 
in  the  jacket. 

If  x  denotes  the  weight  of  steam  liquefied  in  the  jacket  per 
I.  H.  P.  per  hour,  then  the  heat  supplied  by  the  jacket  is  x  times 
the  latent  heat  of  steam  corresponding  to  the  pressure,  plf  of  the 
jacket  steam.  The  total  heat  expended  in  the  performance  of 


BOILER  AND  ENGINE  EFFICIENCY  115 

effective  work  per  I.  H.  P.  per  hour  is  then  (H?  —  H  2  )  23.37  +  xH[. 
This  expenditure  of  heat  is  balanced  by  the  effective  work  done. 
The  gross  work  per  stroke  is  144/?wi;2,  and  the  equivalent  in  work  of 
the  heat  energy  contained  in  the  p2  steam  is  l±4p.2v.,.  The  effective 

,     .    ,,        1  44  v.(  »«—»„")   .       va  i  Pi 

work  per  stroke  is  then  ~      >  m  wnlc"  P*  -         " 


34^ 

27.78  pounds,  and  i>2  is  the  displacement  per  stroke  =  0.368  cubic 
feet.     Since  the  weight  of  steam  used  per  stroke  is  0.02473  pound, 

n  3R8 
there  will  be  A  (30473  =  14.8  cubic  feet  displaced  by  the  piston  for 

each  pound  of  steam  used.     Therefore,  the  effective  work  for  the  ex- 

,000fV  ,      .   ,         .   144X14.8(61.92-27.78)23.37 

penditure  of  23.37  pounds  of  steam  is  —  —  ==•«  — 

thermal  units.     Then  we  shall  have  : 

-  37.78)23.37    wheuce 


x  =  r144  X  14.8  X  34.14  _HT+H  r~| 23.37 
=  (69.52  -  1181  +  1157)  ^7  =  1.832 


pounds. 

T-,™  .  ,         .  Effective  work 

Efficiency  of  engine  =  TT — T- 


84,704  X  23.37 
778 


~  [1181  —  (210  -  32)]  23.37  +  1.832  X  887 

=  0.1015,    the    temperature    of    the    feed 
being  taken  at  210°. 

The  steam  liquefied  in  the  jacket  is  returned  to  the  boiler  at  the 
temperature  of  the  water  in  the  boiler,  and  therefore  the  heat  ex- 
pended in  the  jacket  is  less  than  the  total  heat  of  formation  of 
the  liquefied  steam  by  the  amount  of  heat  required  to  raise  the  feed 
water  to  the  temperature  of  the  boiler  steam.  Hence  for  each 
pound  of  steam  liquefied  in  the  jacket  there  is  an  expenditure  of 
1181  —  (210  —  32)  units.  We  shall  have,  therefore, 

Weight  of  jacket  steam  per  I.  H.P.  per  hour  —  '       ,      ^ 

—  1.42  pounds. 

The  total  weight  of  steam  per  I.  H.  P.  per  hour  is  then 
23.37  +  1.42  =  24.79  pounds. 


116  THE  ELEMENTS  OF  STEAM  ENGINEERING 

If  the  saturation  curve,  BD,  of  Fig.  29,  is  to  be  maintained  dur- 
ing expansion,  heat  must  be  obtained  from  a  jacket  or  from  some 
other  external  source.  Steam  under  ordinary  conditions  of  practice 
contains  moisture,  and  if  the  steam  jacket  supplies  sufficient  heat 
to  evaporate  this  moisture  and  also  to  prevent  liquefaction  during 
expansion,  the  curve  approximates  the  hyperbola  BC,  which  repre- 
sents the  expansion  of  a  perfect  gas  at  constant  temperature. 

It  is  found  that  considerable  differences  in  the  amount  of  heat 
supplied  by  a  jacket  do  not  make  very  appreciable  differences  in  the 
form  of  the  expansion  curve,  which  fact  detracts  from  the  value  of 
an  analysis  of  the  performance  of  a  steam  engine  from  indicator 
diagrams. 

Experiments  have  shown  that  no  particular  advantage  is  derived 
in  having  jacket  steam  of  a  pressure  more  than  a  few  pounds  higher 
than  that  of  the  initial  steam  in  the  cylinder,  so  it  is  the  practice 
to  make  the  jacket  steam  of  the  second  and  succeeding  cylinders 
of  stage  expansion  engines  pass  through  reducing  valves. 

The  conditions  of  hyperbolic  expansion  are  not  infrequently  real- 
ized, and  it  is  the  universal  practice  to  regard  the  expansion  curve 
as  a  hyperbola,  which  greatly  facilitates  calculations. 

130.  Condensation  and  Production  of  Vacuum. — After  using 
steam  in  the  cylinder  of  a  condensing  engine  it  is  exhausted  into  a 
condenser  and  brought  into  contact  with  a  jet  of  water,  as  in  the 
case  of  a  jet  condenser,  or  else  it  comes  in  contact  with  a  series  of 
tubes  through  which  cold  water  is  circulating,  as  in  the  case  of  a 
surface  condenser.  The  result  in  either  case  is  the  almost  instant 
condensation  of  the  steam  and  the  production  of  a  vacuum.  The 
perfection  of  the  vacuum  depends  almost  entirely  upon  the  existing 
temperature  in  the  condenser.  If  the  temperature  were  32°,  the 
corresponding  pressure  would  be  0.085  Ib.  per  square  inch,  and  the 
vacuum  nearly  perfect.  No  such  vacuum  as  this  is  ever  attained  in 
a  condenser,  nor  would  it  be  desirable,  inasmuch  as  there  must 
always  be  an  excess  of  pressure  in  the  condenser  over  that  in  the 
air-pump  barrel  in  order  that  the  valves  should  open. 

With  the  surface  condenser  only  the  water  resulting  from  the 
condensation  of  the  steam  accumulates  in  the  condenser,  but  more 
or  less  air,  due  to  leakage  or  liberated  from  the  feed  water,  is  also 
present,  and  if  it  were  allowed  to  accumulate,  the  vacuum  would 
soon  be  destroyed.  For  this  reason  the  air-pump  is  necessary  with 


BOILER  AND  ENGINE  EFFICIENCY  117 

both  styles  of  condenser,  and  its  function  is  to  remove  from  the  con- 
denser the  air  and  uncondensed  vapor,  as  well  as  the  water  resulting 
from  the  condensation.  The  ordinary  temperature  of  the  water 
after  condensation  is  about  126°,  corresponding  to  a  pressure  of 
2  pounds  per  square  inch. 

-Heat  Rejected  into  the  Condenser. — Steam  is  commonly  exhausted 
into  the  condenser  at  a  pressure  of  from  4  to  5  pounds  with  a  cor- 
responding temperature  of  about  160°.  Assuming  the  temperature 
of  the  water  after  condensation  to  be  126°,  each  pound  of  exhaust 
steam  upon  entering  the  condenser  is  reduced  from  160°  to  126°, 
which  represents  the  liberation  of  160  —  126  =  34  units.  The  lat- 
ent heat  of  a  pound  of  steam  at  160°  temperature  is  1091.7  — 
0.7(160  —  32)  =  1002  units,  so  that  each  pound  of  steam  gives  up 
1002  units  by  its  condensation  into  water  at  160°.  The  total  quan- 
tity liberated  by  the  condensation  of  one  pound  of  the  steam  is 
1002  +  34  =  1036  units,  and  this  heat  must  all  be  destroyed,  or 
abstracted,  by  the  injection  water.  This  quantity  of  heat  is  known 
as  the  heat  rejected  to  the  condenser  per  pound  of  steam  used.  The 
total  heat  of  formation  of  the  steam  will  always  be  greater  than  that 
rejected  to  the  condenser,  and  the  difference,  neglecting  the  small 
loss  by  radiation,  is  the  amount  converted  into  work. 

Weight  of  Condensing  Water  Required  Per  Pound  of  Steam. — 
Suppose,  in  the  case  of  a  jet  condenser,  that  the  temperature  of  the 
injection  water  is  60°,  then,  using  the  figures  given  above,  each 
pound  will  absorb  126  —  60  =  66  units;  therefore,  to  destroy  1036 

i  n^?fi 
units  given  up  by  the  pound  of  steam  will  require  -^-  =  15.7  Ibs. 

of  injection  water. 

In  the  case  of  a  surface  condenser  a  greater  amount  of  water 
will  be  required  for  condensation,  due  to  the  fact  that  the  final  tem- 
perature of  the  injection  water  passing  through  the  tubes  must  be 
lower  than  that  of  the  exhaust  steam;  while  in  the  jet  condenser 
the  injection  water  and  the  condensed  steam  form  a  mixture  at  the 
same  temperature.  Suppose  the  injection  water  after  passing 
through  the  tubes  was  discharged  at  a  temperature  of  85°.  Then 
each  pound  would  absorb  only  85  —  60  =  25  units,  and  there  would 

1  /"kQ/? 

be  required  ~-^-  =  41.44  Ibs.  injection  water  per  pound  of  steam 
condensed. 


118  THE  ELEMENTS  OF  STEAM  ENGINEERING 


PKOBLEMS. 

35.  The  barrel  of  a  calorimeter  contains  175  Ibs.  of  water  at  a 
temperature  of  60°.     Ten  pounds  of  steam  at  a  pressure  of  150  Ibs. 
absolute  are  blown  into  the  barrel,  after  which  the  temperature  of 
the  mixture  is  found  to  be  120°.    The  temperature  of  steam  at  150 
Ibs.  is  358°,  and  the  latent  heat  861.2  units.    Find  the  dryness  frac- 
tion of  the  steam.  Ans.  0.943. 

36.  The  efficiency  of  a  boiler  is  0.7;  steam  pressure,  135  Ibs.  per 
gauge;  temperature  of  the  feed  water,  120° ;   thermal  value  of  the 
coal  used,  14,200  units  per  pound.     Find  the  evaporation  of  satu- 
rated steam  per  pound  of  coal.  Ans.  9.01  Ibs. 

37.  The  efficiency  of  a  boiler  is  0.75;  steam  pressure,  150  Ibs.  per 
gauge;  temperature  of  feed,  130°;   dryness  fraction  of  the  steam, 
0.95 ;  thermal  value  of  the  coal,  14,300  units  per  pound.     Find  the 
number  of  pounds  of  steam  of  the  given  quality  evaporated  per 
pound  of  coal.  Ans.  10.19  Ibs. 

38.  A  boiler  under  certain  conditions  evaporates  9.25  Ibs.  water 
per  pound  of  coal  into  steam  at  364°  from  feed  water  at  70°,  the 
dryness  fraction  of  the  steam  being  0.9.     Under  different  circum- 
stances 8.75  Ibs.  of  water  are  evaporated  per  pound  of  coal  into 
steam  at  380°  from  feed  water  at  102°,  the  fraction  of  moisture  in 
the  steam  being  0.05.     Which  of  the  two  cases  is  the  more  efficient  ? 

Ans,  First  case  is  4.13  per  cent  more  efficient. 

39.  Steam  containing  4  per  cent  of  moisture  is  generated  at  165 
Ibs.  gauge  pressure  from  feed  water  at  152°.     The  evaporation  per 
pound  of  coal  is  8.5  Ibs.     In  another  instance  8  Ibs.  of  water  per 
pound  of  coal  are  evaporated  into  steam  at  125  Ibs.  pressure  from 
feed  water  at  132°,  the  dryness  fraction  being  0.97.     Find  the 
equivalent  evaporations  from  and  at  212°,  and  under  the  supposi- 
tion that  coal  in  the  first  instance  cost  $4.10  per  ton  and  in  the 
second  $3.40  a  ton,  show  which  case  is  the  more  economical. 

Ans.  Equivalent  evaporation  first  case,  9.25  Ibs. 

Equivalent  evaporation  second  case,  8.81  Ibs. 
First  case  14.7  per  cent  more  expensive. 

40.  A  boiler  generates  steam  at  185  Ibs.  gauge  pressure  from  feed 
water  of  temperature  of  152°,  evaporating  9  Ibs.  of  water  per  pound 
of  coal.     The  steam  contains  3  per  cent  of  moisture,  and  the  coal 


BOILER  AND  ENGINE  EFFICIENCY  119 

contains  12  per  cent  of  ash  and  4  per  cent  of  moisture.  Find  the 
actual  evaporation  and  the  equivalent  evaporation  from  and  at  212° 
of  a  pound  of  the  dry  combustible.  Ans.  10.7  Ibs.  and  12  Ibs. 

41.  What  would  be  the  coal  consumption  per  I.  H.  P.  per  hour  in 
the  ideal  engine  working  between  the  temperatures  of  369°   and 
120°,  assuming  the  thermal  value  of  the  coal  to  be  equal  to  that  of 
pure  carbon,  and  the  efficiency  of  the  boiler  to  be  perfect  ? 

Ans.  0.585  Ib. 

42.  How  many  pounds  of  water  must  be  evaporated  per  hour  per 
I.  H.  P.  with  the  ideal  engine,  the  consumption  of  fuel  being  0.585 
Ib.  of  carbon  per  I.  H.  P.  per  hour,  the  temperature  of  the  steam 
369°,  and  that  of  the  feed  water  120°  ?  Ans.  7f  Ibs. 

43.  The  efficiency  of  an  engine  is  14  per  cent,  and  of  the  boiler 
70  per  cent.     The  coal  used  has  a  thermal  value  of  14,300  units 
per  pound.     Find   the   number   of   pounds   of   coal   required   per 
I.  H.  P.  per  hour.  Ans.  1.816  Ibs. 

44.  A  9"  by  10"  engine  uses  steam  at  80  Ibs.  gauge  pressure.   Cut- 
off, 0.25  stroke;  revolutions,  375  per  minute;  clearance,  6  per  cent; 
I.  H.   P.  developed,  46.     The  sensible  heat  and  specific  volume 
of  the  initial  steam  are  294  units  and  4.62  cubic  feet,  respectively. 
The  dryness  fraction  of  the  steam  is  0.97,  the  thermal  value  of  the 
fuel  14,000  units,  the  temperature  of  the  feed  water  182°,  and  the 
efficiency  of  the  boiler  70  per  cent.    Find  (a)  The  pounds  of  steam 
per  I.  H.  P.  per  hour,     (b)  The  pounds  of  water  evaporated  per 
pound  of  coal,     (c)  The  pounds  of  coal  per  I.  H.  P.  per  hour. 

Ans.     (a)  24.17  Ibs.     (&)  9.77  Ibs.     (c)  2.  47  Ibs. 

45.  Diameter  of  cylinder,  14";   stroke,  13";  cut-off,  0.'25  of  the 
stroke;   clearance,  5  per  cent;    initial  absolute  steam  pressure,  95 
Ibs.;  back  pressure,  17  Ibs.;  revolutions  per  minute,  300.     The  spe- 
cific volume  of  steam  at  95  Ibs.  pressure  is  4.62  cubic  feet,  and  it  is 
expected  that  0.9  of  the  theoretical  mean  effective  pressure  will  be 
realized.     Find  the  weight  of  steam  used  per  I.  H.  P.  per  hour. 

Ans.  22.47  Ibs. 

46.  With  a  jet  condenser  the  temperature  of  the  exhaust  steam  is 
193°,  of  the  injection  water  60°,  and  of  the  mixture -in  the  con- 
denser 120°.     Find  the  weight  of  injection  water  per  pound  of  ex- 
haust steam.  Ans.  17.53  Ibs. 


120  THE  ELEMENTS  OF  STEAM  ENGINEERING 

47.  With  a  surface  condenser  the  temperature  of  the  exhaust 
steam  is  170°,  and  of  the  condenser  126°.  The  injection  water 
enters  at  a  temperature  of  60°  and  is  discharged  at  85°.  Find  the 
weight  of  injection  water  per  pound  of  exhaust  steam. 

Ans.  41.56  Ibs. 


CHAPTER  XL 
ENGINE  DESIGN. 

131.  The  Indicator  Diagram  in  Preliminary  Design. — In  the  pre- 
liminary design  of  the  valve  of  an  engine  certain  data  are  assumed, 
from  which  other  dimensions  result  as  a  consequence.  It  is  then 
possible  to  use  the  properties  of  the  indicator  diagram,  regardless 
of  scale,  to  determine  if  the  design  of  the  valve  will  give  satisfac- 
tory results  as  to  mean  pressure  and  steam  consumption,  and  thus 
ascertain  if  the  design  is  satisfactory. 

In  Fig.  31  let: 


L  =  length  of  stroke. 

Jc  =  length  of  stroke  to  cut-off. 

c  =  clearance  in  fraction  of  stroke. 

x  =:  distance  from  end  of  stroke  at  which  exhaust  closes. 
p1  =  absolute  initial  pressure. 
p2  =  absolute  terminal  pressure. 

p3  —  absolute  back  pressure  =.  initial  pressure  of  compression. 
pc  =  absolute  terminal  pressure  of  compression. 
v±  =  k  -\-  c  =  initial  volume. 
v2  =  L  +  c  =  terminal  volume. 

=  r  =  ratio  of  expansion. 
—  rc  —  ratio  of  compression. 


122 


THE  ELEMENTS  OF  STEAM  ENGINEERING 


The  expansion  and  compression  curves  being  hyperbolic,  we  shall 

rn  ry 

have  P^V!  =  p2v2,  whence  —  =  —  =  r\ 

and  in  like  manner  —  —  rc. 

Denote  the  area  HBCDKR  by  A,  the  area  WAHR  by  5,  the  area 
WRKGO  by  C,  and  the  area  GKDO  by  D. 

The  area  A  represents  the  effective  work  and  is  equal  to  pe  X  L, 
from  which  we  have  : 

A      A  +  B+C+D—  (B  +  C+D)   . 


e  1S  tne   mean 


effective  pressure. 


•*r  C   -X 


It  has  been  shown  on  page  99  that  the  area  of  the  whole  diagram 
A  +  B  -\-  C  +  D  is  p-iPid  +  l°ger)?  and  the  area  C  is  likewise 


x).    There- 


- x)] 


Area  of  B  =  c(p±  —  pc),  and  area  of  D  =  pB(L 
fore, 


_ 


The  simplest  way  of  finding  pe  is  to  calculate  the  areas  A,  B,  C, 
and  D  from  the  data  given,  and  substitute  their  values  in  the 
equation  just  found. 

EXAMPLE  I.  —  Initial  pressure  of  steam,  88  Ibs.,  and  terminal 
pressure,  27  Ibs.  Stroke,  30";  clearance,  4  per  cent  of  piston  dis- 
placement. A  terminal  pressure  of  compression  of  60  Ibs.  is  de- 
sired, the  back  pressure  being  17  Ibs.  All  the  pressures  being 
absolute,  it  is  desired  to  find  the  point  of  cut-off,  the  point  of 
exhaust  closure,  and  the  mean  effective  pressure. 


ENGINE  DESIGN  123 

Draw  Fig.  32  without  regard  to  scale.     We  have  r  =  J~f  =  3.26, 
the  loge  of  which  is  1.18. 
c  —  30  X  0.04  =  1.2  inches. 

p1v1  =  p2v2)  or  SS(k  +  1.2)  =  27  X  31.2,  whence  k  =  8.37  inches. 
17  (a  +  1.2)=  60  X  l-25  whence  x  —  3.04  inches. 

rc  —  ^  =  3.53,  or  rc  =  f°-  =  3.53.         loge  rc  =  1.26. 

Area  (A  +  B  +  C  +  D)  —  p^(l  +  loge  r)  =  88  X  9.57  X  2.18 

=  1835.9. 

Area  B  =  c(p1—pc)=  1.2(88  —  60)=  33.6. 
Area  C  =  pcvc(l  +  logc  re)  —  60  X  1.2  X  2.26  =  162.72. 
Area  D  —  p3(L  —  x)  =  17(30  —  3.04)  =458.32. 
Area  A  —  Area  (A  +  B  +  C  +  D)  —  Area  (B  +  C  +  D) 
=  1835.9  —  (33.6  +  162.72  +  458.32)  =  1181.26. 


pe=Are?A  =         '=  39.375  pounds  per  square  inch. 

Jj  60 

If  it  is  determined  that  the  engine  shall  make  170  revolutions  per 
minute  and  that  the  diameter  of  the  cylinder  be  26  inches,  the 
horse-power  to  be  expected  will  be: 

T  TT  T>        pLaN      39.375  X  2  5  X  r  X  169  X  340 
LILR  =  3pOO  =  33,000 

132.  To  estimate  the  steam  consumption  per  I.  H..  P.  per  hour  we 
would  proceed  thus  : 

Referring  to  Fig.  32,  the  volume  of  steam  in  the  cylinder  at 

cut-off  is   ^  ityoQ      cubic  feet,  in  which  k  and  c  are  in  inches,  and 

1  //co 

a  is  the  area  of  the  piston  in  square  inches  ;  and  if  si  be  the  specific 
volume  of  the  steam  at  pressure  pi}  the  weight  of  the  steam  in  the 

cylinder  at  cut-off  is  T,y9UN/       pounds.     The  volume  of  steam  in 

JL  /  /vo  X  »i 

i  cr  -\-  c\ft 

the  cylinder  at  the  moment  of  exhaust  closure  is    v  ^  ^J     cubic 


feet,  and  if  ss  be  the  specific  volume  of  steam  at  pressure  ps,  the 
weight  of  this  steam  is    lo  pounds.     This  last  weight  of 


steam  is  compressed  into  the  clearance  space,  and  only  the  weight 
(k  +  c)a       (x  4-  c)a    .      ,,  .      ,,  ,   . 

1728  X  s  — 1728  X      1S  al^owe^  to  escaPe  in  the  exhaust,  and  is, 

therefore,  the  weight  of  steam  used  per  stroke.     If  R  be  the  number 


124  THE  ELEMENTS  OF  STEAM  ENGINEERING 

of  revolutions  per  minute  of  the  engine,  we  shall  have  :     Pounds  of 
steam  per  I.  H.  P.  per  hour 

["(*  +  g>  _  Qg  +  c>»1g  X  2  X  60       Ra 
_  |_1728  X  g|      1728  X 


I.  H.  P.  14.4  I.  H.  P. 

EXAMPLE.  —  Find  the  number  of  pounds  of  steam  used  per  hour 
per  I.  H.  P.  by  the  engine  of  the  example  on  page  122,  the  specific 
volume  of  steam  at  pressures  of  88  Ibs.  and  17  Ibs.  being  4.9  and 
23.22  cubic  feet,  respectively. 

Here  we  have  : 
Pounds  of  steam  per  I.  H.  P.  per  hour 


170  X  3.1416  X  169  X  201.99 

14.4  X  538.7  '    14.4  X  538.7  X  4.y  X  23.22 

=  21.13  Ibs.  Adding  25  per  cent  (see  Sec.  149,  p.  150),  we  get 
26.4  Ibs.  as  an  approximate  result. 

133.  Size  of  Cylinder  for  a  Given  Power. — The  diameter  of  the 
cylinder  of  an  engine  depends  upon  the  piston  speed  and  the  mean 
pressure  permissible  in  obtaining  the  power  desired. 

The  mean  pressure  depends  upon  the  initial  pressure  and  the 
ratio  of  expansion,  and  may  be  calculated,  under  the  assumption 
that  the  expansion  takes  place  according  to  Boyle's  Law,  by  the 

formula  pm  =    ^      ' — -^— -  .     Or,   the  mean  pressure  may  be 

obtained  very  approximately  by  drawing  an  ideal  diagram  from 
the  conditions  of  the  case,  from  which  the  actual  diagram  to  be 
expected  may  be  drawn  and  the  mean  pressure  obtained. 

The  mean  pressure  derived  theoretically  is  never  realized  in 
practice,  for  there  are  always  causes  which  make  the  mean  pressure 
from  the  actual  indicator  diagram  less  than  that  due  the  initial 
pressure  and  ratio  of  expansion  used  in  the  design. 

The  principal  causes  which  tend  to  make  the  actual  mean  pres- 
sure less  than  that  theoretically  due  the  initial  pressure  and  the 
ratio  of  expansion  are :  Wire-drawing ;  liquefaction  in  the  cylinder ; 
release  of  the  steam  before  the  piston  arrives  at  the  end  of  the 
stroke ;  clearance ;  compression  and  back  pressure ;  friction  in  the 
ports  and  passages ;  and,  in  stage  expansion  engines,  "  drop." 

Attempts  are  sometimes  made  to  treat  the  effects  of  these  causes 
analytically,  and  though  such  treatment  is  interesting  and  to  a 
degree  instructive,  it  involves  so  many  assumptions  which  may  or 
may  not  be  realized  in  practice  that  the  results  obtained  are  of  little 
value. 


ENGINE  DESIGN 

These  causes  have  not,  of  course,  the  same  effect  in  all  cases,  but 
experience  has  shown  that  a  factor,  depending  upon  the  type  and 
the  working  conditions  of  the  engine,  may  be  applied  to  the  theo- 
retical pressure  obtained  in  any  case,  and  the  result  be  taken  as  the 
mean  pressure  to  be  expected. 

There  is  a  substantial  agreement  among  the  authorities  as  to 
these  mean  pressure  factors,  and  they  may  be  taken,  under  the  con- 
ditions named,  as  follows: 

For  high-speed  stationary  engines,  from  0.87  to  0.94;  for  two- 
stage  expansion  engines,  0.75  to  0.85;  for  triple-expansion  engines 
of  the  mercantile  marine,  0.65  to  0.7;  for  triple-expansion  engines 
of  war  vessels,  0.6  to  0.65;  for  torpedo-boat  engines,  0.45  to  0.55. 

134.  Piston  Speed. — The  speed  of  the  piston  is  determined  from 
considerations  of  convenience,  the  type  of  the  engine,  and  the  nature 
of  the  work  to  be  done.     It  depends,  of  course,  upon  the  length  of 
the  stroke  and  the  number  of  revolutions,  and  is  expressed  in  feet 
per  minute.     The  mean  piston  speed  iks  'equal  to  the  product  of 
twice  the  length  of  stroke  in  feet  and  the  number  of  revolutions  per 
minute.     Experience  has  shown  that  a  good  piston  speed  for  slow 
pumping  engines  is  from  125  to  175  feet  per  minute;   for  ordinary 
horizontal  engines,  from  300  to  450  feet;   for  high-speed  stationary 
engines,  from  500  to  650  feet;    for  marine  engines,  from  700  to 
800  feet;  for  locomotives,  from  900  to  1000  feet;  and  for  torpedo- 
boat  engines  a  piston  speed  as  high  as  1200  feet  has  been  attained. 

135.  Stroke. — There  is  no  standard  rule  for  determining  the 
stroke  of  an  engine,  but  the  local  condition  as  to  space  is  often  the 
controlling  influence  in  its  determination.     Strokes  of  stationary 
engines  vary  from  10  inches  to  60  inches,  and  those  of  marine  en- 
gines from  18  inches  to  50  inches. 

136.  Cylinder  Ratios. — There  is  a  lack  of  agreement  among  the 
authorities  as  to  the  proper  ratios  between  the  cylinders  of  stage 
expansion  engines.     It  is  the  aim  to  equalize  the  power  between  the 
cylinders,  but  this  may  be  effected  within  practical  limits  by  means 
of  the  adjustment  of  the  points  of  cut-off,  and  quite  independently 
of  the  cylinder  ratios,  but  a  too  low  cylinder  ratio  with  equality  of 
powers  may  cause  a  pronounced  inequality  in  the  initial  stresses 
on  the  pistons.     The  important  considerations  of  making  the  weight 
of  the  machinery  and  the  space  it  occupies  as  small  as  possible  in 
a  vessel  of  war,  together  with  the  fact  that  engines  of  such  vessels 


126  .  THE  ELEMENTS  OF  STEAM  ENGINEERING 

are  seldom  worked  at  their  maximum  power,  make  it  desirable  that 
their  cylinder  ratios  be  made  smaller  than  those  of  engines  of  the 
mercantile  marine,  where  the  conditions  are  somewhat  reversed. 

The  proposition  has  been  advanced  that,  with  given  initial  and 
terminal  pressures,  percentages  of  clearance,  point  of  cut-off  in  the 
H.  P.  cylinder  and  the  pressure  at  that  point,  the  ratio  of  the  net 
areas  of  the  high-  and  low-pressure  pistons  follows  as  a  consequence. 
The  reasoning  employed  in  the  establishment  of  this  proposition  is 
logical  and  interesting,  but  as  the  operation  is  long  and  tedious, 
and  as  it  involves  several  assumptions  which  can  only  be  justified 
by  a  reference  to  the  results  of  successful  practice,  the  question 
very  naturally  arises,  Why  not  at  once  select  for  any  particular  case 
the  cylinder  ratio  employed  in  successful  practice?  The  construc- 
tion of  the  theoretical  diagram  and  the  investigation  of  the  initial 
stresses  on  the  pistons  would  then  reveal  the  fact  if  the  selected 
ratio  were  improper. 

Successful  practice  has  shown  that  in  the  two-stage  expansion 
engine,  with  the  initial  pressure  ranging  from  80  Ibs.  to  120  Ibs., 
the  ratio  of  the  low-pressure  cylinder  to  the  high-pressure  cylinder 
should  vary  from  3  to  4.5  for  engines  of  the  mercantile  marine,  and 
from  2.5  to  4  for  engines  of  vessels  of  war.  For  triple-expansion 
engines  of  the  mercantile  marine,  with  pressures  ranging  from 
160  Ibs.  to  180  Ibs.,  the  ratio  of  the  low-pressure  cylinder  to  the 
high-pressure  cylinder  varies  from  6  to  7.5.  For  engines  of  war 
vessels,  where  the  pressures  range  from  160  Ibs.  to  250  Ibs.,  the 
ratio  varies  from  5  to  7. 

Opinions  are  at  variance  concerning  the  proper  size  of  the  inter- 
mediate cylinder  of  a  triple-expansion  engine.  To  make  it  a  mean 
between  the  H.  P.  and  the  L.  P.  cylinders,  which  is  good  practice, 
its  area  of  cross-section  would  be  the  square  root  of  the  product 
of  the  cross-sectional  areas  of  the  H.  P.  and  L.  P.  cylinders.  Let 
d  and  D  denote  the  diameters  of  the  H.  P.  and  L.  P.  cylinders 
respectively,  and  let  x  denote  the  diameter  of  the  intermediate 
cylinder.  Tf  <p  denotes  the  ratio  between  the  L.  P.  and  H.  P. 
cylinders,  then 


„    ,  *Dn-      ,  M      D°-     „,.       ,         ,       D1  D 

But  —r-  =  —IT  >  whence  d1  =  — .    Therefore  x1  =  —/= ,  and  x  —  -77=-. 

44'  <p  <*  (> 


That  is,  the  diameter  of  the  intermediate  cylinder  is  equal  to  the 


ENGINE  DESIGN  127 

diameter  of  the  low-pressure  cylinder  divided  by  the  fourth  root 
of  the  ratio  of  the  low-pressure  cylinder  to  the  high-pressure 
cylinder. 

EXAMPLE  I. — Eequired  the  dimensions  of  an  engine  to  develop 
46  horse-power;  piston  speed,  625  feet  per  minute;  absolute  pres- 
sure, 95  Ibs. ;  cut-off,  0.25  stroke;  back  pressure,  2  Ibs.  above  the 
atmosphere;  clearance,  5  per  cent. 

Here  we  have : 

Eatio  of  expansion  =  --,         =  n  05  ^_  Q  QK  —  3.5,  the  loge  of  which 

is  1.2528. 
Theoretical  mean  pressure  =  95(1  +  *-2628)  =  61.12  Ibs. 

Theoretical  mean  effective  pressure  =  61.12  —  16.7  =  44.42  Ibs. 
Using  a  mean  pressure  factor  of  0.9,  we  shall  have : 
Expected  mean  effective  pressure  =  44.42  X  0.9  =  39.978,  say  40 
pounds. 

40  X  625  X  3.1416  d* 


,,         /  46  X  33,000  X  4  .  .     , 

=  V  40X625X3.1416  =  8'8>  Say  9  mcheS' 
The  piston  speed  is  the  product  of  twice  the  stroke  in  feet  and  the 
number  of  revolutions  per  minute ;  so  if  we  assume  the  stroke  to  be 
10  inches,  we  shall  have : 

-D      T    , .                                 625  X  12 
Eevolutions  per  minute  =  — ^ =  375. 

EXAMPLE  II. — Eequired  the  dimensions  of  a  compound  engine  to 
develop  2000  I.  H.  P.;  piston  speed,  700  feet  per  minute;  absolute 
initial  pressure,  112  Ibs.;  back  pressure  in  the  condenser,  3  Ibs.; 
cut-off  in  high-pressure  cylinder,  f  stroke;  clearance  of  high-pres- 
sure cylinder,  14  per  cent,  and  of  low-pressure  cylinder,  12  per  cent. 

As  the  engine  is  for  the  mercantile  marine,  a  cylinder  ratio  of 
4.25  and  a  mean  pressure  factor  of  0.8  will  be  chosen. 

Since  the  L.  P.  cylinder  is  the  measure  of  the  power  of  a  stage 
expansion  engine,  its  diameter  is  first  determined  and  made  suffi- 
ciently large  to  develop  the  whole  of  the  power  in  that  cylinder  from 
the  mean  pressure  derived  from  the  expansion.  The  whole  power  is 
then  apportioned  equally  between  the  cylinders. 


Ratio  of  expansion  =  -  =  "  +  =  9.24,  the  log 

of  which  is  2.224. 


128  THE  ELEMENTS  OF  STEAM  ENGINEERING 

112(1  +  2.224) 
Theoretical  mean  pressure  =  ^  ~ .        -  =  39.08  Ibs. 

Theoretical  mean  effective  pressure  =  39.08  —  3  =  36.08. 

Mean  effective  pressure  to  be  expected  =  36.08  X  0.8  =  28.864  Ibs. 

pLaN    ,,  33,000  X  2000 

L  H'  R  =  3pOO>  therefore>  a  =  28.864  X  700   r=  3266'5 
inches. 


/  QO£«  O    K 

Diameter  of  L.  P.  cylinder  =    J         '.    =  64.49,  say  64.5  inches. 


K 

Area  of  H.  P.  piston  =    ^     '    =  768.6  square  inches. 


Diameter  of  H.  P.  cylinder  =      $Tf$fo=  31.283,  say  31.25  inches. 
Assuming  the  stroke  as  3  feet,  the  engine  would  then  have  to 

iy(\(\ 

make  Q       0  =117  revolutions  per  minute. 

o  X  <6 

EXAMPLE  III.  —  Find  the  dimensions  of  a  triple-expansion  engine 
to  be  used  in  a  ship  of  the  mercantile  marine,  the  horse-power  to  be 
developed  being  3600.  Piston  speed,  800  feet  per  minute;  initial 
absolute  pressure,  155  Ibs.  ;  absolute  back  pressure,  3  Ibs.  ;  cut-off 
in  H.  P.  cylinder  at  0.6  stroke.  The  estimated  clearances  are  18 
per  cent  for  the  H.  P.  cylinder  and  15  per  cent  for  the  L.  P. 
cylinder. 

Assume  a  ratio  of  7  between  the  volumes  of  the  L.  P.  and  H.  P. 
cylinders,  and  a  mean  pressure  factor  of  0.7,  such  assumptions 
being  in  accordance  with  good  practice.  Then, 

r  =  7(1  +  O-15)  =  10.32,  the  loge  of  which  is  2.337. 
U.o  +.v/«lo 

Theoretical  mean  pressure  =  --  10  32*          "  ^'^  ^s' 
Theoretical  mean  effective  pressure  =  50.12  —  3  =  47.12  Ibs. 
Mean  effective  pressure  to  be  expected  =  47.12  X  0.7  =  32.984, 

say  33  Ibs. 
I.  H.  P.=  S*£  Before,  B  =  "ffOXMOO  =  «00  square 

inches. 

Diameter  of  L.  P.  cylinder  =  ,y/^f^  =  ?5-69>  saJ  75-75  inches- 
Diameter  of  the  I.  P.  cylinder  =  =  46-535>  saJ  46-5  inches. 


Diameter  of  the  H.  P.  cylinder  =  y  7  x  Ut7854  =  28.61,  say  28f 
inches. 


ENGINE  DESIGN 

Assuming  a  stroke  of  30  inches,  the  engine  would  have  to  make 

800 
2  fr  x  2  ~  -^  revolutions  per  minute. 

If  triple-expansion  engines  of  3600  horse-power  were  required 
for  a  vessel  of  war,  a  smaller  cylinder  ratio,  a  smaller  mean  pressure 
factor,  and  a  longer  H.  P.  cut-off  would  have  been  assumed  in 
order  to  meet  the  changed  conditions,  thus:* 

EXAMPLE  IV.  —  Find  the  dimensions  of  a  triple-expansion  engine 
of  3600  horse-power  to  be  used  in  a  vessel  of  war.  Piston  speed, 
800  feet  per  minute;  initial  absolute  steam  pressure,  155  Ibs.; 
absolute  back  pressure,  3  Ibs.;  cut-off  in  H.  P.  cylinder,  0.725 
stroke.  The  estimated  clearances  are,  18  per  cent  for  the  H.  P. 
cylinder  and  15  per  cent  for  the  L.  P.  cylinder. 

Assume  a  ratio  of  5.25  between  the  volumes  of  the  L.  P.  and  the 
H.  P.  cylinders,  and  a  mean  pressure  factor  of  0.6.  Then  : 


r  =    Q5  +  o  ffi  =  6'66'  the  loge  °f  which  is  L898' 

„,,  -         155(1  +  1.898) 

Theoretical  mean  pressure  =         v  „  „„  —    •*  =  67.43  Ibs. 

Theoretical  mean  effective  pressure  =  67.43  —  3  =  64.43  Ibs. 
Mean  effective  pressure  to  be  expected  =  64.43  X  0.6  =  38.658. 

pLaN    ,,  33,000  X  3600 

L  H"  P'==  3boO'thereforeo=      38.658  X  800      =  3841-4  sq. 
inches. 

Diameter  of  L.  P.  cylinder  = 

69  93 

Diameter  of  I.  P.  cylinder  =    4  JLJ_  =  46.2,  say  46.25  inches. 

*y  5./c5 

Diameter  of  H.  P.  cylinder  =  \/5  25^  0  7854  =  30-875>  saJ  31  in* 

137.  Steam  per  I.  H.  P.  per  Hour.—  From  the  data  of  Example  I, 
Art.  136,  the  initial  volume  of  steam  per  stroke,  expressed  in  stroke. 
is  10  X  0.25  +  10  X  0.05  =  3  inches.  Then, 

Steam  used  per  hour  =  8x8.1416X45X^x875x8X60  ^  ^ 

1  /  Ao 

cubic  feet. 

The  specific  volume  of  steam  at  95  pounds  pressure  is  4.62  cubic 
feet,  hence  : 

4-Q70 
Steam  per  I.  H.  P.  per  hour  =  4>6a  ^  46  =  23.39  Ibs. 

9 


130  THE  ELEMENTS  OF  STEAM  ENGINEERING 

It  would  probably  be  safe  to  increase  this  by  5  per  cent,  making 
the  steam  per  I.  H.  P.  per  hour  24.56  Ibs. 

138.  Coal  per  I.  H.  P.  per  Hour. — Assuming,  in  the  example  just 
considered,  that  the  boiler  evaporates  8  pounds  cf  water  per  pound 

of  coal,  we  would  then  have,  — ^—  =3.07  pounds  of  coal  expended 
per  I.  H.  P.  per  hour! 

139.  The  steam  space  in  a  boiler  depends  upon  the  volume  of 
steam  used  by  the  engine  and  not  upon  the  weight  of  steam  used. 
Two  boilers  having  the  same  grate  and  heating  surface  will  evapo- 
rate the  same  weight  of  steam  in  a  given  time,  but  if  the  pressures 
differ  the  volumes  of  the  steam  will  differ.     Slow-running  engines 
with  comparatively  low  steam  pressures  will,  therefore,  require  boil- 
ers having  larger  steam  space  per  unit  of  power  than  are  necessary 
for  engines  of  higher  speed  and  higher  pressures  using  the  same 
weight  of  steam.     There  should  be  allowed  0.8  of  a  cubic  foot  of 
steam  space  per  I.  H.  P.  for  slow-running  pumping  engines.    This 
allowance  is  reduced  as  the  character  of  the  engines  calls  for  higher 
speeds  and  higher  pressures,  until  a  minimum,  probably,  of  0.35 
cubic  foot  per  I.  H.  P.  is  reached. 

For  the  high-speed  engine  of  Example  I,  Art  136,  an  allowance 
of  0.5  cubic  foot  of  steam  space  per  I.  H.  P.  would  be  sufficient.  The 
steam  space  of  the  boiler  would  then  be  46  X  0.5  =  23  cubic  feet. 

The  steam  used  per  minute  by  the  engine  would  be  -g^-  =  82.83 
cubic  feet.  The  boiler  would,  therefore,  be  emptied  of  steam 
?%P  =  3.6  times  a  minute,  giving  the  steam  JE-gte  16.67  seconds 

/CO  «5.O 

to  dry. 

140.  Steam  Port  Dimensions. — In  determining  the  proper  area  of 
steam  ports  the  controlling  influences  are  that  the  exhaust  must  be 
free,  thus  minimizing  the  back  pressure,  and  that  excessive  clearance 
must  be  avoided.     Almost  invariably  the  exhaust  steam  from  the 
cylinder  must  pass  through  the  same  port  through  which  it  was 
admitted,  and  for  that  reason  the  size  of  the  port  must  be  governed 
by  the  velocity  it  is  desirable  the  exhaust  steam  should  have.     A 
mean  velocity  for  the  exhaust  of  6000  feet  per  minute  is  in  accord- 
ance with  good  practice,  so  that  the  area  of  the  ports  through  the 
valve  should  be  calculated  from  that  basis.     As  will  be  seen  later, 
the  port  is  never  completely  open  for  the  admission  of  steam, 


ENGINE  DESIGN  131 

though  the  opening  should  not  be  so  contracted  as  to  give  to  the 
admission  steam  a  velocity  exceeding  12,000  feet  per  minute. 

It  is  evident  that  for  the  free  escape  of  the  exhaust  steam  the 
relation,  area  of  piston  X  velocity  of  piston  =  area  of  port  X  ve- 
locity of  steam,  must  obtain.  From  this  we  have  : 

Area  of  steam  port  =  area  of  piston  ^velocity  of  piston  _ 

We  shall  have,  then,  for  the  engine  of  Example  I,  Art.  136  : 

A          -    ,  3.1416  X  4.5  X  4.5  X  625        a  co  .         .     , 

Area  of  steam  port  =  -  -  =  6.624  sq.  inches. 


Taking  the  length  of  the  port  as  0.8  the  diameter  of  the  cylinder 
we  shall  have  : 

Width  of  port  =    6'624    =  0.92,  say  jf  inch. 

141.  Mechanical  Advantage  of  the  Stage  Expansion  Engine.  — 

To  illustrate  the  mechanical  advantage  of  the  stage  expansion 
engine  in  producing  a  reduction  in  the  initial  stress,  and  a  decrease 
in  the  ratio  of  initial  to  mean  stress,  over  that  produced  with  the 
simple  expansive  engine,  a  comparison  will  be  made  between  a  sim- 
ple expansive  engine,  with  two  cylinders,  each  having  an  area  A/2, 
and  a  compound  engine  of  two  cylinders,  the  ratio  of  whose  piston 
areas  is  3.  The  area  of  the  L.  P.  piston  will  then  be  A,  and  that  of 
the  H.  P.  piston,  A/3.  In  each  case  the  initial  pressure  will  be  80 
Ibs.  absolute,  the  ratio  of  expansion  6,  and  the  back  pressure  in  the 
condenser  4  Ibs.  The  receiver  pressure  of  the  compound  engine  will 
be  taken  as  22  Ibs.  Each  engine  has  the  same  number  of  working 
parts,  but  a  separate  expansion  valve  would  be  required  on  each  of 
the  cylinders  of  the  simple  engine  to  effect  so  early  a  cut-off  as,  at 
least,  J  stroke. 

(  1  )   The  simple  expansive  engine  with  two  cylinders  : 

Mean  pressure  in  each  cylinder  =  -  (T^  =  37.22  Ibs. 

4 

Effective  initial  load  on  each  piston  =  (80  —  4)^-  =  38  A  Ibs. 

Effective  mean  load  on  both  pistons  —  33.22  X  A  Ibs. 

Efficiency  of  the  svstem      Mean  effec.  load  obtained  =  33MA 

Mean  effec.  load  expected      33.22A 

(2)   The  two-cylinder  compound  engine: 

If  the  volume  of  the  H.  P.  cylinder  be  denoted  by  unity,  that  of 
the  L.  P.  cylinder  must  be  denoted  by  3,  and  if  v^  denotes  the  vol- 
ume of  the  H.  P.  cylinder  up  to  cut-off,  we  shall  have  : 


132  THE  ELEMENTS  OF  STEAM  ENGINEERING 

3  3 

-  =  r,  or   -  =  6,  whence  v±  =  0.5  cut-off  in  H.  P.  cylinder.     The 

ratio  of  expansion  in  the  H.  P.  cylinder  is,  therefore,  TT-=  =  2. 

u.o 

To  insure  a  uniform  pressure  in  the  receiver,,  the  product  of  the 
volume  and  pressure  given  to  it  by  the  H.  P.  cylinder  must  equal  the 
product  of  the  volume  and  pressure  taken  from  it,  and  we  shall 
have  80  X  0.5  =  22  X  3  X  &,  whence  Tc  =  0.606  cut-off  in  L.  P. 
cylinder  to  maintain  a  receiver  pressure  of  22  Ibs. 

The  ratio  of  expansion  in  L.  P.  cylinder  =  rA    —  1-65. 


Mean  pressure  in  H.  P.  cylinder  =  80(1  +^°'6931)  _  67>72  lbs> 

,,  22(1  +  0.5008) 

Mean  pressure  in  L.  P.  cylinder  =  -         ..  „„  --  -  =  20.03  Ibs. 

Mean  effective  pressure  in  H.  P.  cylinder  =  67.72  —  22  =.  45.72  Ibs. 
Mean  effective  pressure  in  L.  P.  cylinder  =.  20.03  :  —  4  =  16.03  Ibs. 

Effective  initial  load  on  H.  P.  piston  =  (80  —  22)  ^  —  19.331  Ibs. 

Effective  initial  load  on  L.  P.  piston  =(22  —  4)  A  —  181  Ibs. 
Mean  effective  load  obtained  on  both  pistons 

=  45.72  X  4  +  16.031  =  31.271  Ibs. 
o 

Mean  pressure  due  an  initial  pressure  of  80  Ibs.  and  a  ratio  of 

expansion  of  6,  is  80(1  +g1-7918)  =  37.32  Ibs. 
Mean  effective  load  expected  =  (37.22  —  4)1  =  33.221  Ibs. 
Efficiency  =  =0.941. 


In  comparing  the  results  it  is  found  that  the  initial  load  on  each 
of  the  pistons  of  the  simple  engine  is  381  Ibs.,  while  that  on  the 
H.  P.  piston  of  the  compound  engine  is  19.381  Ibs.,  and  on  the 
L.  P.  piston  181  Ibs.,  showing  that  the  initial  loads  on  the  pistons 
of  the  compound  engine  are  practically  equal,  and  their  aggregate 
less  than  that  on  either  of  the  pistons  of  the  simple  engine. 

The  ratio  of  the  mean  load  to  the  initial  load  on  each  piston  of 

33.22  X^ 

the  simple  engine  is      ao  .       =  ±^  =-0.437,  or  43.7  per  cent. 

OoA  on 

The  same  ratio  for  the  pistons  of  the  compound  engine  are  as  fol- 
lows: 


ENGINE  DESIGN  133 

45.72  X  4 
On  H.  P.  piston  =      19  33^    =  °-788>  or  78-8  Per  cent- 


On  L.  P.  piston  =  =  0.89,  or  89  per  cent. 


In  consequence  of  this  reduction  in  the  initial  and  mean  loads, 
the  various  parts  of  the  compound  engine  whose  size  depends  on 
these  factors  may  be  made  lighter  than  those  of  the  simple  engine, 
and  the  friction  on  the  guides  and  on  the  bearings  will  be  much  less 
severe.  The  ratios  of  the  mean  to  the  initial  loads  being  much 
nearer  unity  in  the  case  of  the  compound  engine  than  with  the 
simple  engine,  the  turning  moment  is  much  more  uniform  and  the 
engine  will  run  more  steadily.  The  question  of  uniform  load  and 
steadiness  in  running  is  one  of  great  importance  in  marine  practice, 
but  with  stationary  engines  where  the  use  of  the  fly  wheel  is  permis- 
sible its  importance  is  lessened. 

The  mean  effective  loads  on  the  pistons  of  the  compound  engines 
show  the  work  to  be  very  evenly  divided  between  the  cylinders, 
but  the  total  work  falls  short,  of  that  of  the  simple  engine  by 
1  —  0.941  =  0.059,  or  5.9  per  cent.  This,  loss  may  be  attributed 
to  "  drop,"  which,  in  this  case,  is  -8/  —  22  =  18  Ibs. 

A  consideration  of  the  range  of  temperatures  in  the  cylinders  will 
show  a  very  marked  advantage  in  favor  of  the  compound  engine. 

The  temperature  of  steam  at  80  Ibs.  pressure  is  312°,  and  that  of 
steam  at  4  Ibs.  pressure  is  153°.  The  range  of  temperature  is  then 
312  —  153  =  159°  in  each  cylinder  of  the  simple  engine. 

The  temperature  of  steam  at  22  Ibs.  pressure  is  233°,  therefore 
312  —  233  =  79°  is  the  range  of  temperature  in  the  H.  P.  cylinder 
of  the  compound  engine,  and  233  —  153  =  80°  is  the  range  of  tem- 
perature in  the  L.  P.  cylinder.  It  is  thus  seen  that  the  range  of 
temperature  in  each  cylinder  of  the  compound  engine  is  just  about 
one-half  that  in  the  cylinders  of  the  simple  engine.  The  advan- 
tages which  would  accrue  from  this  distribution  of  temperature  are 
set  forth  in  Art.  58,  p.  38. 

142.  Crank-pin  -Piston  Velocity  Diagram.  —  In  some  problems 
connected  with  steam  engine  design,  it  is  required  to  know  the 
velocity  of  the  piston  when  the  velocity  of  the  crank-pin  is  known  ; 
this  can  very  readily  be  done. 


134 


THE  ELEMENTS  OF  STEAM  ENGINEERING 


Let  (7P,  Fig.  33,  be  the  position  of  the  crank  when  the  connecting- 
rod  PQ  makes  an  angle  a  with  the  center  line  of  the  engine.  Pro- 
duce the  connecting-rod  until  it  intersects  the  vertical  through  C 
at  R.  The  direction  of  P  at  any  point  is  in  the  tangent  at  the  point ; 
and  if  we  denote  the  velocity  of  the  crank-pin  by  V,  and  that  of  the 


piston  by  v,  we  shall  have  the  resolved  velocities  of  P  and  Q  along 
PQ  equal,  since  PQ  is  a  rigid  rod. 
Hence  : 

V  sin  <f>  =.  v  cos  a. 
Consequently 

—  —  s*n^  _      sinp      _  CR 
V  "  5c¥a  .   sin  CRQ  ~~  ~UP ' 


Now  if,  to  any  scale,  OP  denotes  the  assumed  constant  velocity  of 
the  crank-pin,  then  CR  will,  to  the  same  scale,  denote  the  velocity  of 
the  piston. 

We  may  construct  a  curve  of  piston  velocities,  as  follows : 

Let  CP  and  PQ,  Fig.  34,  be  the  positions,  respectively,  of  the 

crank  and  connecting-rod,  corresponding  to  the  piston  position  Q. 

Produce  QP  until  it  intersects  the  vertical  through  C  at  R.     With 

C  as  a  center  and  CR  as  a  radius,  describe  the  arc  Rr,  intersecting 


ENGINE  DESIGN  135 

CP  at  r;  then  r  is  a  point  of  the  curve.  The  locus  of  r  is  the 
polar  curve  of  piston  velocities,  the  radius  vector  in  line  of  the 
crank  representing  the  piston  velocity  for  that  crank  position,  to 
the  same  scale  that  the  length  of  the  crank  represents  the  velocity 
of  the  crank-pin.  If  the  connecting-rod  were  of  infinite  length,  the 
locus  of  r  would  be  two  circles  described  on  the  vertical  positions  of 
the  crank  as  diameters. 

The  velocities  of  the  crank-pin  and  piston  are  equal  when  the 
crank  is  in  its  vertical  positions,  and  again  at  crank  positions  Ct 
and  Ct',  where  the  connecting-rod  produced  passes  through  8  and  8'. 
Between  the  points  8  and  t,  and  8'  and  tr,  the  velocity  of  the  piston 
is  greater  than  that  of  the  crank,  and  at  all  other  points  it  is  less. 

The  piston  has  its  maximum  velocity  very  nearly  when  the  crank 
and  connecting-rod  are  at  right  angles,  or  when  the  connecting-rod 
is  tangent  to  the  crank-pin  circle. 

PEOBLEMS. 

48.  Stroke  of  engine,  24  inches;  initial  absolute  pressure  of  steam 
at  cylinder,  100  Ibs.,  but  wire-drawing  reduces  the  pressure  to  95  Ibs. 
at  cut-off.     Back  pressure,  18  Ibs.;   cut-off,  0.25  stroke;   clearance, 
10  per  cent.    Eelease  takes  place  at  90  per  cent  of  stroke,  and  ex- 
haust closes  after  75  per  cent  of  the  return  stroke  is  completed. 
The  expansion  and  compression  being  hyperbolic,  construct  the  ex- 
pected indicator  diagram,  and  find  the  terminal  pressure,  p2;   the 
final  pressure  of  compression,  pc;   the  mean  effective  pressure,  pe, 
by  means  of  ordinates;   and  the  estimated  steam  consumption  per 
I.  H.  P.  per  hour. 

Ans.  £2  =  31.8;   pc  =  63;   pe  —  42.8;.  steam 
per  I.  H.  P.  per  hour  —  22.47  Ibs. 

49.  Stroke  of  engine,  20";  diameter  of  cylinder,  18";  revolutions 
per  minute,  180;   clearance,  5  per  cent;   terminal  pressure,  28  Ibs.; 
back  pressure,  17  Ibs.;    ratio  of  expansion,  3.5;   ratio  of  compres- 
sion, 4.     The  specific  volumes  of  steam  at  the  initial  and  back  pres- 
sures are  4.44  cubic  feet  and  23  cubic  feet  respectively.     Using  the 
properties  of  the  indicator  diagram,  it  is  required  to  find  the  mean 
effective  pressure,  and  assuming  that  85  per  cent  of  it  will  be  real- 
ized, find  the  I.  H.  P.,  and  the  number  of  pounds  of  steam  used  per 
I.  H.  P.  per  hour. 

Ans.  pe  —  42.17;  I.  H.  P.  =  165.73;  steam  per  I.  H.  P. 
per  hour,  22.6  Ibs. 


136  THE  ELEMENTS  OF  STEAM  ENGINEERING 

50.  An  18.5"  x  30"  engine  makes  129  revolutions  per  minute,  the 
consumption  of  fuel  being  5.5  long  tons  a  day.     The  mean  effective 
pressure  is  36  Ibs.,  and  26  pounds  of  steam  are  used  per  I.  H.  P. 
per  hour.     Find  the  number  of  pounds  of  coal  burned  per  I.  H.  P. 
per  hour,  and  the  number  of  pounds  of  water  the  boiler  must  evapo- 
rate per  pound  of  coal. 

Ans.  2.73  Ibs.  of  coal  per  I.  H.  P.  per  hour,  and  9.524  Ibs. 
of  water  evaporated  per  pound  of  coal. 

51.  An  engine,  18.5"  x  30",  makes  129  revolutions  per  minute. 
The  mean  effective  pressure  is  36.6  Ibs.  per  square  inch,  and  the 
consumption  of  coal  per  I.  H.  P.  per  hour  is  3  Ibs.     The  weight  of 
steam  used  per  stroke  is  0.3478  Ib.     Find  the  number  of  pounds  of 
steam  used  per  I.  H.  P.  per  hour;  the  coal  consumption  (long  tons) 
per  day;   the  number  of  pounds  of  water  the  boiler  must  evaporate 
per  pound  of  coal. 

Ans.  Pounds  of  steam  per  I.  H.  P.  per  hour,  28 ;  tons  of 
coal  per  day,  6.18 ;  pounds  of  water  to  be  evapo- 
rated per  pound  of  coal,  9.33  Ibs. 

52.  Find  the  cylinder  dimensions  of  an  engine  to  develop   75 
horse-power.     Piston  speed,  600  feet  per  minute ;  absolute  pressure 
of  steam,  95  Ibs.;   cut-off,  0.3  stroke;   back  pressure,  2  Ibs.  above 
the  atmosphere;  clearance,  5  per  cent. 

53.  Required  the  dimensions  of  a  compound  engine  for  use  in  the 
mercantile  marine  to  develop  1800  I.  H.  P.     Piston  speed,  725  feet 
per  minute;  absolute  initial  pressure,  112  Ibs.;  back  pressure  in  the 
condenser,  2  Ibs. ;   cut-off  in  H.  P.  cylinder,  0.4  stroke.    Estimated 
clearances:  in  H.  P.  cylinder,  10  per  cent;  in  L.  P.  cylinder,  12 
per  cent. 

54.  Find  the  dimensions  of  a  triple-expansion  engine  to  be  used 
in  the  mercantile  marine,  the  horse-power  to  be  developed  being 
4000.     Piston  speed,  800  feet  per  minute ;  initial  absolute  pressure, 
160  Ibs.;  absolute  back  pressure,  3  Ibs.;   cut-off  in  H.  P.  cylinder, 
0.6  stroke.     The  estimated  clearances  are:    16  per  cent  for  the 
H.  P.  cylinder  and  14  per  cent  for  the  L.  P.  cylinder. 

55.  What  steam-port  dimensions  should  be  given  to  an  engine 
designed  to  have  a  piston  speed  of  800  feet  a  minute,  the  diameter 
of  the  cylinder  being  15  inches? 


CHAPTEE  XII. 
THE  ZEUNER  VALVE  DIAGRAM. 

143.  The  essential  features  and  functions  of  the  slide  valve  have 
already  been  considered,  and  it  is  now  the  intention  to  consider, 
with  a  degree  of  particularity  sufficient  for  practical  purposes,  the 
main  features  of  its  design. 

No  detail  of  the  steam  engine  is  of  more  importance  than  the 
correct  design  of  its  valve  gear,  which  includes  the  valve  and  the 
mechanism  that  gives  it  motion.  A  defective  design  may  occasion 
a  loss  in  fuel  as  great  as  20  per  cent,  and  cause  an  unevenness  in 
the  motion  of  the  engine,  with  a  resulting  wear  and  tear  in  the 
working  parts  that  may  lead  to  serious  casualty. 

The  angularities  of  the  connecting  and  eccentric  rods  introduce 
irregularities  into  the  motion  of  the  piston  and  of  the  valve.  Mathe- 
matical formula  showing  their  relative  positions  during  the  dis- 
tribution of  the  steam  in  the  cylinder  are  too  abstruse  and  com- 
plicated for  practical  use,  but  graphic  methods  have  been  de- 
vised to  simplify  the  problem,  of  which  that  of  Zeuner  is 
as  simple  and  complete  as  any.  It  permits,  without  difficulty, 
the  angularity  of  the  connecting-rod  to  be  taken  into  account, 
though  not  that  of  the  eccentric  rod;  but  since  the  length  of  the 
eccentric  rod  is  great  compared  with  the  throw  of  the  eccentric,  its 
angularity  introduces  an  inappreciable  irregularity,  and  the  motion 
given  to  the  valve  is  assumed  to  be  harmonic.  The  rule  and  com- 
passes are  the  only  instruments  needed  in  the  construction  of  this 
diagram,  and  its  accuracy  depends  largely  upon  the  skill  of  the 
draftsman. 

144.  Take  AB,  Fig.  35,  equal  to  the  travel  of  the  valve,  and  bisect 
it  at  0.     Let  0  represent  the  center  of  the  shaft,  OA  the  direction 
of  the  crank  when  on  the  dead  center,  and  OE  the  direction  of  the 
center  line  of  the  eccentric,  the  angular  advance  being  denoted  by  0. 
On  AB  as  a  diameter  describe  a  circle.     The  intersection,  E,  of 
this  circle  with  OE  will  represent  the  center  of  the  eccentric,  since 
the  half-travel  of  the  valve  is  equal  to  the  distance  between  the 
center  of  the  shaft  and  the  center  of  the  eccentric. 


138 


THE  ELEMENTS  OF  STEAM  ENGINEERING 


The  motion  of  the  valve  being  assumed  harmonic,  the  perpen- 
dicular, EC,  let  fall  from  E  upon  AB,  gives  00  as  the  distance  the 
valve  has  moved  from  its  mid-position  (see  Fig.  16).  As  the  shaft 
rotates  the  crank  will  assume  the  directions  Oa,  Oa',  etc.,  and  the 
eccentric  the  positions  Oe,  Oe',  etc.;  and  if  from  e  and  e'  perpen- 
diculars be  let  fall  upon  AB,  the  distances  Of  and  Of  intercepted 
between  the  center  of  the  circle  and  the  feet  of  the  perpendiculars 
will  represent  the  distances  the  valve  has  moved  from  its  central 
position  when  the  crank  has  the  directions  Oa  and  Oa'. 

Instead  of  the  crank  and  eccentric  moving  in  their  circular  paths, 
suppose  them  to  remain  fixed  and  the  radius  OB  to  revolve  in  the 


opposite  direction  and  assume  the  positions  06  and  06',  corres- 
ponding to  Oa  and  Oa'.  From  E  let  fall  perpendiculars  Es,  ES* 
on  Ob  and  06'.  By  similar  triangles  it  is  easily  shown  that  05,  Osf 
are  equal  to  Of,  Of',  therefore  Os,  Os'  represent  equally  well  the 
movements  of  the  valve  from  its  central  position  when  the  crank 
assumes  the  directions  Oa,  Oa'.  Since  the  angles  OsE,  Os'E  are 
right  angles,  the  locus  of  the  points  s,  s'  is  the  circumference  of  a 
circle  described  on  OE  as  a  diameter. 

If,  then,  upon  OE  as  a  diameter  a  circle  be  described,  the  radii 
vectors  Os,  Os'  give  the  distances  the  valve  has  moved  from  its 
central  position  for  the  angular  positions  06,  06'  of  the  crank,  it 
being  remembered  that  for  the  eccentric  positions  Oe,  Oe'  the  real 
crank  positions  would  be  Oa,  Oa'. 


THE  ZEUNER  VALVE  DIAGRAM 


139 


This  artifice  is  usual  and  convenient  in  the  construction  of  the 
Zeuner  diagram  and  contributes  to  its  simplicity. 

A  precisely  similar  construction  and  reasoning  will  show  that  a 
circle  described  on  a  diameter  OE',  equal  to  and  diametrically  oppo- 
site OW,  so  that  EE'  is  a  straight  line,  will  give,  by  means  of  the 
crank  intercepts,  the  movement  of  the  valve  during  the  return 
stroke. 

We  now  have  a  diagram  which  shows  the  distance  the  valve  has 
moved  from  its  central  position  for  any  position  of  the  crank,  and 
therefore  the  corresponding  position  of  the  piston  may  be  found. 

145.  The  angular  advance  of  the  eccentric  may  be  obtained  as 
follows : 

Describe  a  circle,  AEBE',  Fig.  36,  with  the  throw  of  the  eccen- 
tric as  a  radius.  Let  OA  be  the  direction  of  the  crank  when  on 


36 


the  dead  center.  From  0  set  off  in  a  direction  away  from  the  crank 
the  distance  OC  equal  to  the  lap  plus  the  lead.  Draw  the  vertical 
CE,  cutting  the  circle  at  E.  Then  OE  will  be  the  position  of  the 
eccentric  radius  when  the  crank  is  on  the  dead  center,  and  the  angle 
DOE  will  be  the  angular  advance  of  the  eccentric.  This  follows 
from  the  fact  that  the  valve  has  moved  from  its  central  position  a 
distance  equal  to  the  lap  plus  the  lead  when  the  crank  is  on  the 
dead  center.  The  above  construction  for  the  angular  advance  holds 
when  the  valve  takes  steam  at  its  outside  edges;  but  if,  as  is  not 
infrequently  the  case,  the  valve  takes  steam  at  its  inside  edges,  the 
lap  plus  the  lead  must  be  laid  off  from  0  towards  the  crank,  and 
OE'  will  then  be  the  position  of  the  eccentric,  and  DOE'  the  angle 
of  advance. 

146.  With  a  radius  equal  to  the  throw  of  the  eccentric,  or  the  half- 
travel  of  the  valve,  describe  the  circle  AEC',  Fig.  37.     Let  OA  be 


140 


THE  ELEMENTS  OF  STEAM  ENGINEERING 


the  direction  of  the  crank  when  on  the  dead  center,  and  OE  the 
direction  of,  and  equal  to,  the  throw  of  the  eccentric.  Draw  the 
diameter  DD'  perpendicular  to  OA.  From  E  drop  a  perpendicular 
to  AB,  intersecting  it  at  N.  ON  is  the  distance  the  valve  has 
moved  from  its  mid-position  when  the  crank  is  on  the  dead  center. 
From  0  set  off  ON'  equal  to  ON,  and  draw  N'C  perpendicular  to- 
AB.  Draw  the  diameter  CO'.  The  angle  COD  —  angle  DOE  = 
angular  advance  of  the  eccentric.  If  the  crank  move  to  Oa,  the 
eccentric  will  take  the  position  Oe,  such  that  the  angles  AOa  and 
EOe  are  equal.  Drop  ef  perpendicular  to  AB.  Then  Of  will  be 
the  distance  the  valve  has  moved  from  its  central  position  for  the 


crank  position  Oa.  On  Oa  set  off  Of  —  Of.  From  what  has  been 
shown  in  Fig.  35,  all  points  such  as  N'9  f  will  lie  in  the  circum- 
ferences of  two  circles  described  on  the  diameters  OC  and  OC't 
which  are  each  equal  to  the  half  travel  of  the  valve,  and  make  an 
angle  with  the  perpendicular  to  the  line  of  dead  centers  equal  to 
the  angular  advance  of  the  eccentric,  measured  in  a  direction  oppo- 
site to  that  of  the  motion  of  the  crank. 

If,  with  0  as  a  center  and  a  radius  OL  equal  to  the  outside  lap 
of  the  valve,  a  circle  be  described,  the  distance  the  port  is  open  for 
the  admission  of  steam  from  any  crank  position,  as  Oa,  is  given  by 
the  intercepted  part  rf.  This  will  be  well  understood  from  the 
fact  that  the  opening  of  the  port  is  equal  to  the  movement  of  the 
valve  from  its  central  position,  minus  the  lap. 

Since  ON'  is  the  movement  of  the  valve  from  its  central  position 
when  the  crank  is  on  the  dead  center,  A,  it  must  be  equal  to  the 


THE  ZEUNER  VALVE.  DIAGRAM 


141 


lap  plus  the  lead,  and  since  OL  is  the  lap,  N'L  must,  therefore,  be 
the  lead. 

We  may  now  construct  a  complete  diagram  and,  in  order  that 
it  be  not  too  complicated,  it  will  be  drawn  for  the  steam  and 
•exhaust  on  only  one  side  of  the  piston,  the  side  remote  from  the 
shaft,  and  the  lap,  the  lead,  etc.,  determined  by  the  diagram  will 
be  for  the  end  of  the  valve  remote  from  the  shaft. 

The  stroke  of  the  piston  from  the  head  end  to  the  crank  end  of 
the  cylinder  is  known  as  the  forward  stroke  (top  stroke  in  vertical 
engines),  and  from  the  crank  to  the  head  end,  the  return  stroke 
(bottom  stroke  in  vertical  engines).  The  diagram  will,  therefore, 
give  the  necessary  dimensions  of  the  end  of  the  valve  to  obtain  the 


38. 


admission  and  cut-off  of  the  steam  for  the  forward  stroke,  and  the 
release  and  compression  of  the  same  steam  on  the  return  stroke. 

With  a  center  0,  Fig.  38,  and  radius  OA  equal  to  the  half-travel 
of  the  valve,  describe  a  circle.  A3  being  the  line  of  dead  centers, 
the  diameter  DD'  is  drawn  perpendicular  to  it,  and  the  angle  DOE, 
equal  to  the  angular  advance,  is  laid  off  from  DO  in  a  direction 
contrary  to  that  of  the  motion  of  the  crank  as  indicated  by  the 
arrow.  On  the  radii  OE  and  OE'  of  the  diameter  EEf  as  diam- 
eters, describe  the  circles  OyE  and  OXE'.  For  the  stroke  under 
consideration  the  first  of  these  circles  is  known  as  the  primary  valve 
circle,  or  steam  circle,  and  the  second  as  the  secondary  valve  circle, 
or  exhaust  circle. 


14:2  THE  ELEMENTS  OF  STEAM  ENGINEERING 

With  a  radius  OL  equal  to  the  steam  lap  of  the  valve,  describe 
the  arc  sLsf.  Now,  since  the  admission  and  cut-off  of  the  steam 
must  take  place  when  the  distance  of  the  valve  from  its  central 
position  is  equal  to  the  lap,  it  follows  that  s  and  s',  the  intersec- 
tions of  the  lap  circle  with  the  steam  circle,  give  the  crank  positions 
OF  and  00  for  admission  and  cut-off  respectively.  In  like  man- 
ner, release  and  compression  occur  when  the  distance  of  the  valve 
from  its  central  position  is  equal  to  the  inside  lap.  If,  then,  with 
0  as  a  center  and  a  radius  OL'  equal  to  the  inside  lap,  an  arc  be 
described,  its  intersections  e  and  e'  with  the  exhaust  circle  will  give 
the  crank  positions  OQ  and  OQ'  for  release  and  compression  respec- 
tively. 

Should  the  exhaust  lap  be  negative,  which  is  not  infrequently 
the  case  for  the  top  stroke  of  vertical  engines,  the  intersections  of 
the  exhaust  lap  circle  with  the  steam  circle  must  be  taken  to  get 
the  crank  positions  for  release  and  compression. 

This  will  become  evident  after  a  consideration  of  the  fact  that 
the  intercept  on  the  crank  position  by  the  steam  or  the  exhaust  circle 
gives,  in  every  instance,  the  distance  of  the  valve  from  its  central 
position.  The  valve  is,  therefore,  centered  when  the  intercept  is 
zero,  that  is,  when  the  crank  is  in  the  position  OP  or  OP',  tangent 
to  both  circles;  PP'  is,  therefore,  perpendicular  to  EE'.  Now,  if 
the  exhaust  lap  is  positive,  the  crank  will  have  to  revolve  beyond 
OP  before  the  port  opens  to  release,  and  will  be  intersected  by  the 
exhaust  circle.  If,  however,  the  exhaust  lap  be  negative,  release 
will  have  taken  place  before  the  valve  is  centered,  or  before  the 
crank  reaches  the  position  OP,  and  the  intercept  will,  therefore,  be 
made  by  the  steam  circle. 

For  any  position  of  the  crank,  as  OF,  Oy  is  the  distance  of  the 
valve  from  its  central  position,  and  the  intercepted  portion  xy 
between  the  lap  and  the  steam  circles  is  the  amount  the  port  is  open 
for  the  admission  of  steam.  The  intercepts  in  the  hatched  area 
of  the  steam  circle  show  port  openings  to  steam,  and  therefore  LG 
is  the  measure  of  the  lead. 

From  the  relative  positions  of  the  crank  and  eccentric  and  the 
harmonic  motion  of  the  valve,  it  is  seen  that  from  the  crank  posi- 
tions OF  to  OE  the  opening  of  the  port  to  steam  continually  in- 
creases, at  first  quickly  and  finally  slowly;  from  OE  to  00  the 
valve  is  closing  the  port,  at  first  slowly  and  finally  quickly. 


THE  ZEUNER  VALVE  DIAGRAM  143 

The  intercepts  in  the  hatched  area  of  the  exhaust  circle  show 
openings  of  the  port  to  exhaust.  The  port  will,  of  course,  be  wide 
open  to  exhaust  when  the  valve  has  moved  from  its  central  position 
a  distance  equal  to  the  exhaust  lap  plus  the  width  of  the  port;  and 
any  further  distance  the  throw  of  the  eccentric  may  cause  the  valve 
to  move  will  result  in  overtravel.  If  with  0  as  a  center  and  a 
radius  equal  to  the  exhaust  lap  plus  the  width  of  the  port,  an  arc 
XZ  be  drawn,  the  port  will  be  shown  wide  open  to  exhaust  during 
the  period  from  X  to  Z,  and  E'E"  will  show  the  overtravel. 

To  draw  the  diagram  for  the  return  stroke,  the  exhaust  circle 
for  the  forward  stroke  becomes  the  steam  circle  for  the  return 
stroke,  and  vice  versa. 

If  the  steam  and  exhaust  laps  are  the  same  for  both  strokes,  the 
lap  circles  must  be  completed  so  as  to  intersect  the  opposite  circles, 
and  thus  give  the  corresponding  events  of  the  return  stroke.  Fre- 
quently, however,  they  are  different  for  the  two  strokes,  in  which 
event  the  correct  radii  must  be  used. 

This  is  essentially  the  Zeuner  valve  diagram,  and  by  its  applica- 
tion the  numerous  problems  of  the  slide  valve  may  be  solved.  By 
varying  some  of  the  points,  the  alterations  in  the  others  may  easily 
be  found ;  and  by  assuming  certain  elements  the  remainder  may  be 
determined. 

Some  confusion  exists  at  times  as  to  what  is  meant  by  the  differ- 
ent names  given  to  the  strokes  of  the  piston.  It  should  be  remem- 
bered that  the  stroke  known  as  -forward,  outward,  top,  head,  or 
down,  is  that  in  which  the  piston  moves  toward  the  shaft;  and  the 
stroke  known  as  return,  inward,  bottom,  crank,  or  up,  is  that  in 
which  the  piston  moves  away  from  the  shaft. 

147.  There  are  a  number  of  geometrical  features  of  the  diagram 
which  are  very  useful  in  the  solution  of  problems. 

1.  The  line  joining  the  points  of  admission  and  cut-off  is  tan- 
gent to  the  lap  circle. 

In  Fig.  39  let  OF  and  00  be  the  crank  positions,  respectively, 
for  admission  and  cut-off,  and  let  N'LN  be  an  arc  of  the  lap  circle. 
Join  PC,  cutting  EE'  at  some  point  M.  It  is  required  to  prove 
that  M  is  the  point  of  tangency  of  PC  and  the  arc  N'LN.  The 
triangle  FOG  is  isosceles,  and,  since  ME  is  the  greatest  opening  of 
the  port,  it  is  midway  between  the  admission  and  cut-off  of  the 
steam;  therefore  EEf  bisects  the  vertical  angle  FOC,  and  is  per- 


144 


THE  ELEMENTS  OF  STEAM  ENGINEERING 


pendicular  to  the  base  FC  and  bisects  it  at  M..  Draw  EN,  and  in 
the  right  triangles  EON  and  COM,  we  shall  have  the  hypotenuse 
CO  equal  to  the  hypotenuse  EO.  The  acute  angles  at  E  and  at  C 
are  equal  because  their  sides  are  respectively  perpendicular  to  each 
other ;  therefore  the  triangles  are  equal,  and  ON  =  OM.  But  ON 
is  the  lap ;  therefore  if,  with  0  as  a  center  and  ON  as  a  radius,  the 
arc  N'LN  be  described  it  will  be  tangent  to  FC  at  M . 

A  similar  construction  and  reasoning  will  show  that  a  line  join- 
ing the  point  of  release,  Q,  with  the  point  of  exhaust  closure,  or  of 
compression,  Qf,  will  be  tangent  to  the  exhaust  lap  circle. 

2.  If,  from  the  dead  point  as  a  center,  an  arc  be  described  that 
will  be  tangent  to  the  line  joining  the  points  of  admission  and 
cut-off,  the  radius  of  this  circle  will  be  equal  to  the  lead. 


Fiy.  39 


With  dead  point  A,  Fig.  39,  as  a  center,  describe  an  arc  tangent 
to  FC',  then  its  radius  AT  will  be  equal  to  the  lead.  Through  A 
draw  A8  parallel  to  FC,  and  therefore  perpendicular  to  EE'.  The 
right  triangles  ASO  and  EGO  are  equal,  and  GO  —  80.  But 
LO  =  MO,  therefore  GL  =  8M  =  AT  =  the  lead. 

3.  If  the  points  of  admission  and  cut-off,  and  the  maximum 
opening  of  the  port  to  steam  were  given,  or  if  the  points  of  release 
and  compression  and  the  width  of  port  be  given  (taking  the  maxi- 
mum opening  of  port  to  exhaust  as  the  width  of  port),  there  would 
not  be  sufficient  data  to  construct  the  diagram.  We  can,  however, 
determine  from  the  data  given,  in  either  case,  the  travel  of  the  valve 
and  then  construct  the  diagram. 


THE  ZEUNER  VALVE  DIAGRAM 


145 


With  center  0,  Fig.  40,  describe  the  indefinite  circle  EQQ'.  Let 
Q  and  Q'  be  the  points,  respectively,  of  release  and  compression. 
These  points  would  be  given  by  the  angles  BOQ  and  AOQf,  or  by 
some  definite  part  of  the  stroke  of  the  piston,  now  represented  in- 


Fi'q. 

definitely  by  AB.  Join  QQ',  and  from  what  has  preceded  we  know 
that  QQ'  will  be  tangent  to  the  exhaust  lap  circle,  and  that  EE',  the 
perpendicular  to  QQ'  through  0,  will  give  the  position  of  the  eccen- 
tric, and,  indefinitely,  the  diameters  of  the  steam  and  exhaust  cir- 
cles. WE'  will  then  represent  to  some  scale  the  maximum  opening 
to  exhaust,  which  is  assumed  to  be  "the  width  of  the  port,  given  in 
this  case  as  1.5  inches.  We  would  then  have  the  proportion : 

Travel  of  Valve  _  EE'  _  EE'        Travel      2.5 

Width  of  Port     "  WE' 


The  practical  method  of  determining  the  travel  would  be  as  fol- 
lows: 

Draw  ab  and  ac,  Fig.  41,  making  any  convenient  angle  with 
each  other.  Lay  off  ad  =  WE'  of  Fig.  40  =  ||  inch  by  actual 
10 


146  THE  ELEMENTS  OF  STEAM  ENGINEERING 

measurement.  This  represents  the  width  of  the  port,  in  this  case 
1.5  inches.  Lay  off  ae  =  1.5  inches,  and  join  de.  Our  scale  is 
now  complete.  Lay  off  on  ac  the  distance  of  =  EEr  of  Fig.  40  =  2.5 
inches  by  measurement.,  and  it  will  represent  the  travel  of  the  valve. 

Draw  fg  parallel  to  de  ;  then  ag  =  4  inches  =  travel  of  the  valve. 
The  diagram  can  now  be  constructed. 

If  the  point  of  admission  and  cut-off,  F  and  (7,  Fig.  40,  and  WE 
the  maximum  opening  of  the  port  to  steam  were  given,  the  travel  of 
the  valve  could  be  determined  in  a  similar  manner. 

Travel  of  Valve  EE' 

have>  HSTD^iS^^teSSTtot  ~  WE  ' 


Should  the  point  of  admission,  point  of  cut-off,  and  the  lead  be 
given,  the  travel  could  be  found  from  the  proportion 
Travel  of  Valve      EE' 


Given  Lead 

4.  Since  EGO,  Fig.  39,  is  a  semi-circle,  and  0  one  extremity  of 
its  diameter,  if  at  the  extremity,  G,  of  the  chord,  OG,  (equal  to  the 
lap  plus  the  lead)  a  perpendicular  be  erected,  it  will  meet  the  cir- 
cumferenc  at  E,  the  other  extremity  of  the  diameter. 

148.  An  examination  of  Fig.  38  reveals  12  salient  points  of  in- 
formation, viz.: 

1.  Admission  at  F. 

2.  Cut-off  at  C. 

3.  Eeleaseat§. 

4.  Compression  at  Q'. 

5.  Steam  Lap,  OL. 

6.  Exhaust  Lap,  OL'. 

7.  Steam  Lead,  LG  =  A  T. 

8.  Exhaust  Lead,  L'G'. 

9.  Travel  of  Valve,  EE'. 

10.  Maximum  Opening  of  Port  to  Steam,  EE. 

11.  Maximum  Opening  of  Port  to  Exhaust,  R'E'. 

12.  Angular  Advance  of  the  Eccentric,  DOE. 

Points  1,  2,  >nd  4  are  frequently  given  in  angular  units,  and 
these,  with  some  other  point  in  linear  units,  are  sufficient  to  con- 
struct the  diagram.  These  points  may  be  given  as  occurring  at 
certain  points  of  the  stroke,  and  in  order  to  find  the  corresponding 
crank  positions  the  circle  described  on  EE'  (equal  travel  of  the 
valve)  as  a  diameter  may  be  taken  to  represent  the  travel  of  the 


Given 
Connecting  root,   *4O 
Crvnk  ,  /O. 


for 

of  the   H  P.  Cy/.  o-f  o  triple  -X- 
/fbsolute  steom  pressure.  195 1bs. 


Steom  Jap,  -top  H  ;  bottom 
Exhaust  tap,  1  'op  ^  j 


Travel,3%". 

Mean  cut-off,  7O %. 


top. 


Found 'from 


91  " 

opening,  fop  g^',  bottom 
Ex.    open/'nq,  Ful/  port. 
"' 


Release  "top  0nd  '  bottom,  fa  from  end 


148  THE  ELEMENTS  OF  STEAM  ENGINEERING 

valve  on  a  new  scale;  or  another  circle  described  with  0  as  a 
center,  and  to  any  convenient  scale,  may  be  taken  as  the  crank 
circle. 

Points  1,  2,  5,  7,  and  10  belong  to  the  steam  side  of  the  diagram, 
and  points  3,  4,  6,  8,  and  11  to  the  exhaust  side,  points  9  and  12 
being  common  to  both. 

Three  points  are  sufficient  data  to  construct  either  the  steam 
side  or  the  exhaust  side  alone,  but  for  the  complete  diagram  four 
points  must  be  given,  and  one  or  two  of  these  must  belong  to  a 
different  side  than  the  others.  One  of  the  given  points  must  be  in 
linear  units,  for  if  only  angles  were  given  the  linear  dimensions  of 
the  valve  might  be  made  anything  we  please,  so  long  as  they  bear 
a  certain  ratio  to  each  other. 

If  the  width  of  the  port  in  the  face  of  the  cylinder  be  given  as  a 
part  of  the  data,  it  may  be  taken  as  the  maximum  opening  of  the 
port  to  exhaust,  since  the  opening  to  exhaust  will  be  greater  than 
the  opening  to  steam,  and  the  port  is  made  only  sufficiently  wide 
to  allow  a  full  opening. 

If  the  cut-off  be  one  of  the  given  points  it  is  understood  to  be 
the  mean  cut-off.  The  obliquity  of  the  connecting-rod  occasions  a 
longer  cut-off  on  the  stroke  toward  the  shaft  than  on  the  stroke 
away  from  it,  and  a  compromise  must  be  made  to  secure  the  de- 
sired mean  cut-off. 

Figure  42  is  a  complete  diagram  for  the  valve  of  the  high-pressure 
cylinder  of  a  triple-expansion  marine  engine.  The  theoretical  in- 
dicator diagrams  are  shown  dotted  and  the  diagrams  that  may 
reasonably  be  expected  appear  in  full  lines. 

PEOBLEMS. 

56.  Travel  of  valve,  4.25  inches;  lead  angle,  6°;    angle  at  cut- 
off, 105°.     Find  the  lap,  lead,  and  the  angular  advance  of  the 
eccentric.     Scale,  full  size. 

Ans.  Lap,  1  A  inches;  lead,T\  inch;  angular  advance,  40°  30'. 

57.  Stroke  of  engine,  3  feet;   length  of  connecting-rod,  5  feet; 
steam  lap,  1  inch;   exhaust  lap,  J  inch;   width  of  port,  2  inches; 
overtravel  of  the  exhaust  port,  -J-  inch.     Compression  begins  when 
the  piston  is  one-ninth  of  the  stroke  from  the  end.     Find  the  travel 
of  the  valve,  angular  advance,  lead,  and  the  point  of  cut-off  ex- 


THE  ZEUNER  VALVE  DIAGRAM  149 

pressed  as  a  decimal  of  the  stroke.     Scale  of  stroke,  -J,  and  of  dia- 
gram, full  size. 

Ans.  Travel,  4.75  inches;  lead,  ^  inch;  angular  advance, 
8°;  cut-off,  84.5  per  cent. 

58.  Crank  angle  at  cut-off,  105°;    lead  angle,  8°;    maximum 
port  opening  to  steam,  £  inch.     Find  the  travel  of  the  valve,  the 
lap,  the  lead,  and  the  angular  advance. 

Ans.  Travel,  3.875  inches;  lap,  1  ^  inches;  lead,  -g^-inch; 
angular  advance,  42°. 

59.  Cut-off,  0.7  stroke;   lap,  1  inch;   lead,  -^  inch.     Show  that 
the  angular  advance  is  36°  25'  28",  and  that  the  travel  of  the  valve 
is  4  inches. 

60.  Travel  of  valve,  5  inches;  width  of  port,  2  inches;  opening 
of  port  to  steam,  1.25  inches;  exhaust  lap,  f  inch;  lead,  -f$  inch. 
Find  the  angular  advance,  lap,  point  of  cut-off,  point  of  release, 
point  of  compression,  and  overtravel. 

Ans.  Angular  advance,  35°  6';  lap,  1.25  inches;  cut-off, 
0.71;  release,  94.5  per  cent;  compression,  86  per 
cent;  overtravel,  -J  inch. 


CHAPTEB  XIII. 
ECONOMY  OF  THE  STEAM  ENGINE  AND  BOILER. 

149.  Steam  per  I.  H.  P.  per  Hour  from  Actual  Indicator  Dia- 
grams.— One  of  the  important  uses  of  the  indicator  diagram  is 
to  determine  from  it  the  consumption  of  steam  per  unit  of  power, 
and  thus  get  a  measure  of  the  economy  of  the  engine. 

The  volume  of  steam  in  the  cylinder  at  any  point  of  the  stroke 
is  equal  to  the  displacement  of  the  piston  at  that  point,  plus  the 
volume  of  the  clearance  space.  If  this  volume  be  taken  at  cut-off, 
or  at  any  point  after  cut-off,  and  be  divided  by  the  specific  volume 
of  steam  at  the  absolute  pressure  at  the  point  considered,  the  quo- 
tient will  be  the  weight  in  pounds ;  deducting  from  this  the  weight 
of  the  steam  compressed  at  each  stroke,  we  get  the  weight  of  steam 
supplied  to  the  cylinder  per  stroke — measured  at  the  point  con- 
sidered, and  as  shown  by  the  indicator  diagram.  If  the  weight  of 
steam  per  stroke  be  multiplied  by  the  number  of  strokes  per  hour, 
and  the  product  be  divided  by  the  horse-power  of  the  engine,  the 
quotient  will  be  the  number  of  pounds  of  water  used  per  I.  H.  P. 
per  hour.  The  pressure  at  the  point  considered  is  found  from  the 
diagram,  and  the  volume  per  pound  may  be  taken  from  a  table  of 
the  properties  of  saturated  steam,  or  it  may  be  calculated  by  means 
of  the  formula  pv&  =  482. 

The  consumption  of  steam  per  I.  H.  P.,  obtained  as  above  ex- 
plained, is  based  on  the  weight  of  steam  shown  by  the  indicator 
diagram  to  have  left  the  cylinder  in  the  exhaust,  but  it  does  not 
include  the  waste  of  steam  due  to  liquefaction  in  the  cylinder, 
dampness  of  the  steam,  radiation,  and  other  causes  of  loss.  The 
result  is  valuable  as  a  means  of  comparing  one  engine  with  another, 
or  the  performances  of  the  same  engine  under  different  conditions. 

If  dry  saturated  steam  were  used,  and  if  there  were  no  lique- 
faction during  expansion,  the  quantity  calculated  from  the  diagram 
would  represent  with  approximate  accuracy  the  weight  of  steam 
used.  The  conditions  of  actual  practice  are  such  that  from  20 
to  30  per  cent  should  be  added  to  obtain  results  which  are  approxi- 
mately correct. 


ECONOMY  OF  THE  STEAM  ENGINE  AND  BOILER          151 

Since  there  is  always  water  in  the  cylinder  at  the  point  of  cut-off, 
and  since  more  or  less  of  this  water  is  re-evaporated  into  steam  as 
the  piston  advances,  the  quantity  of  steam  in  the  cylinder  varies 
during  the  stroke.  During  the  exhaust,  too,  some  re-evaporation 
takes  place.  For  these  reasons,  the  consumption  of  steam  shown 
by  the  diagram  is  less  than  the  quantity  actually  passing  through 
the  cylinder.  In  stage  expansion  engines  the  steam  used  in  the 
low-pressure  cylinder  first  passes  through  the  high-pressure  and 
intermediate  cylinders,  and  consequently  the  steam  consumption  of 
the  high-pressure  cylinder  will  be  the  measure  of  consumption  of 
the  whole  engine.  If  the  measure  be  taken  also  from  the  diagrams 
of  the  intermediate  and  low-pressure  cylinders,  as  it  should  be  for 
purposes  of  comparison,  it  will  be  found  that  the  results  from  the 
high-pressure  diagram  will  be  the  greatest,  and  that  the  differences 
between  the  cylinders  may  be  regarded  as  fair  measures  of  the  loss 
in  transmission. 

In  computations  concerning  consumption  of  steam  by  an  engine 
in  any  given  time,  the  volume  of  the  cylinder  and  the  number  of 
revolutions  are,  of  course,  factors;  and  as  these  same  factors  ap- 
pear in  the  power  developed  by  the  engine,  it  follows  that  they 
need  not  appear  in  computing  the  steam  consumption  per  unit  of 
power.  By  reason  of  this  fact  a  constant  quantity,  13,750,  has 
been  determined,  the  employment  of  which  greatly  simplifies  cal- 
culations, enabling  us  to  ascertain  the  steam  consumption  per 
I.  H.  P.  of  an  engine  from  the  indicator  diagram  alone,  and  quite 
independently  of  the  size  and  speed  of  the  engine. 
-  The  use  of  this  constant  is  based  on  the  consideration  that,  if  a 
piston  one  inch  square  in  area  move  12  inches,  the  displacement 

will  be  12  cubic  inches,  or  ^^  of  a  cubic  foot,  and  the  work  done 

will  be  one  foot-pound  for  each  pound  pressure  per  square  inch. 

Since  there  are  33,000  X  60  foot-pounds  in  a  horse-power  con- 
tinuously developed  for  an  hour,  it  follows  that  there  must  be  a 


piston  displacement  of      ?4    6°  =  13,750  cubic  feet  per  I.  H.  P. 

per  hour  when  the  mean  effective  pressure  is  unity,  regardless  of  the 
size  or  speed  of  the  engine;  and  since  the  volume  of  steam  per 
I.  H.  P.  is  inversely  as  the  mean  effective  pressure,  the  quotient  of 


»  n  displacement  in  cubic  feet  per  I.  H.  P. 


152 


THE  ELEMENTS  OF  STEAM  ENGINEERING 


when  the  mean  effective  pressure  is  pe  pounds.  Knowing  the  spe- 
cific volume  of  steam  at  pressure  pc,  its  reciprocal  will  be  the  weight 
of  a  cubic  foot,  and  denoting  this  weight  by  w,  we  shall  have 

13,750  X  w 

— - —     -  as  the  number  of  pounds  of  steam  per  I.  H.  P.  per 

hour,  exclusive  of  losses  from  liquefaction,,  leakage,  etc.,  and  mak- 
ing no  allowance  for  clearance,  compression,  and  expansion. 

The  constant  13,750  is  the  number  of  cubic  feet  of  steam  to  be 
displaced  by  an  engine  in  the  development  of  one  I.  H.  P.  per  hour, 
the  pressure  being  one  pound  per  square  inch,  and  there  being 
neither  clearance,  compression,  release,  nor  expansion. 

Engines,  of  course,  are  never  worked  under  such  conditions,  but 
the  conditions  of  actual  practice  may  be  reduced  to  make  the  con- 


+JD6* 


stant  13,750  applicable  to  the  determination  of  the  steam  consump- 
tion per  I.  H.  P.  per  hour  of  any  engine,  as  shown  by  the  indicator 
diagram,  if  the  diagram  be  given  and  its  scale  be  known. 

In  computing  the  steam  consumption  from  the  indicator  diagram 
it  will  be  well  to  make  the  calculations  both  at  cut-off  and  at 
release,  choosing  points  on  the  expansion  curve  as  close  as  possible 
to  the  points  of  cut-off  and  of  release,  consistent  with  the  conditions 
that  the  port  be  closed  to  steam  in  the  one  case  and  not  yet  open  to 
exhaust  in  the  other.  Should  there  be  no  liquefaction  during  ex- 
pansion the  two  quantities  would  be  the  same. 

Figure  43  is  an  indicator  diagram  from  a  high-speed  engine,  the 
scale  of  the  diagram  being  40,  and  the  clearance  of  the  engine 
6  per  cent. 


ECONOMY  OF  THE  STEAM  ENGINE  AND  BOILER          153 

To  ascertain  the  steam  consumption  per  I.  H.  P.  per  hour,  we 
would  proceed  as  follows  : 

Determine  the  actual  point  of  cut-off,  x,  as  accurately  as  possible. 
The  scale  of  the  diagram  being  40,  draw  the  perfect  vacuum  line, 
00',  14.7  Ibs.  below  the  atmospheric  line.  Erect  perpendiculars  to 
the  atmospheric  line,  touching  the  extremities  of  the  diagram  and 
cutting  the  perfect  vacuum  line  at  A  and  0  '.  By  means  of  a  plani- 
meter,  or  by  means  of  ordinates,  the  mean  effective  pressure  is  found 
to  be  42.4  pounds. 

Then  40  4"  =  ^e  vo^ume  ^n  cubic  feet  to  be  displaced  per  I.  H.  P. 
per  hour,  and  as  this  is  perfectly  independent  of  the  volume  of  the 
cylinder  we  may  assume  the  stroke  volume  AO'  as  the  unit  volume 
of  a  cubic  foot.  Lay  off  the  clearance  volume  AO  =  0.06  to  the 
same  scale  that  AO'  equals  unity.  The  part  of  the  stroke  A'x,  in- 
cluding clearance,  up  to  the  point  of  cut-off  is  found  to  be  0.413  of 
AO',  and  is  the  initial  volume,  v±.  The  volume  of  steam  remaining 
in  the  cylinder  and  clearance  at  exhaust  closure  is  NK  =  0.3125  of 
AO'.  These  volumes  are  proportionally  the  same  as  those  in  the 
actual  cylinder. 

The  pressure  at  cut-off,  x,  and  at  exhaust  closure,  K,  are  found 
by  measurement  to  be  84.5  Ibs.  and  17  Ibs.,  respectively,  and  the 
specific  volumes  of  steam  at  these  pressures  are  5.153  cubic  feet  and 
23.22  cubic  feet.  The  weights  per  cubic  foot  are,  therefore, 

-        =  0.194  Ib.  and  =  0.04307  Ib.  respectively. 


Then,  0.413  X  0.194  —  0.3125  X  0.04307  is  the  weight  of  steam 
sent  to  the  exhaust  per  cubic  foot  displaced,  hence 

10   7Kfk 

•%er;r  (°-413  X  0.194  —  0.3125  X  0.04307)=  21.62  pounds 
of  steam  consumed  per  I.  H.  P  per  hour,  when  measured  at  cut-off, 
and  the  questions  of  clearance,  expansion,  and  compression  having 
been  considered. 

This  result  might  have  been  obtained  by  producing  the  compres- 
sion curve  by  the  eye  until  it  intersects  A'x  at  P".  Then  A'P"  is 
the  volume  of  the  cushion  steam  when  compressed  to  the  pressure  of 
the  steam  at  cut-off,  so  that  P"x  =  0.342  of  AO',  is  the  volume  of 
steam  sent  to  the  exhaust. 

m,         13,750  X  0.342  X  0.194 

Then,  -  ^  ^         -  =  21.62  Ibs.  of  steam  consumed 

per  I.  H.  P.  per  hour,  as  before. 


154  THE  ELEMENTS  OF  STEAM  ENGINEERING 

To  make  the  measurement  at  release,  we  select  a  point  Q  in  the 
expansion  curve  just  before  the  valve  opened  for  release,  and  find 
the  pressure  there  to  be  32  pounds,  the  specific  volume  at  that  pres- 
sure being  12.78  cubic  feet.  The  weight  per  cubic  foot  is  then 

..  a  yg  =  0.07821  Ib.     QE  is  found  by  measurement  to  be  just  equal 

to  AO',  hence  ^|^(1  +  0.07821  —  0.3125  X  0.04307)  =  21  Ibs. 

steam  consumed  per  I.  H.  P.  per  hour  when  measured  at  release, 
indicating  that  21.62  —  21  =  0.62  Ib.  had  liquefied  during  expan- 
sion. 

The  measurement  at  release  might  have  been  made  thus : 
QE  intersects  the  compression  curve  at  P,  and  EP  represents  the 
cushion  steam  when  compressed  to  the  pressure  at  release;  hence 
QP9  which  is  found  by  measurement  to  be  0.83  of  AO',  is  the  vol- 
ume of  steam  sent  to  the  exhaust  at  the  pressure  of  the  release  steam. 

13,750  X  0.07821  X  0.83 

Then,  —  -   =  21.05  Ibs.  steam  consumed  per  . 

42.4 

I.  H.  P.  per  hour  when  measured  at  release,  as  before. 

The  method  frequently  used,  and  the  one  that  gives  results  more 
nearly  correct,  is  as  follows : 

Having  determined  the  point  of  cut-off,  x,  we  find,  by  a  property 
of  the  hyperbola  already  explained,  the  point  c  where  the  cut-off 
would  have  taken  place  to  produce  the  same  expansion  had  there 
been  no  wire-drawing.  Construct  the  theoretical  expansion  and 
compression  curves,  cQ'  and  KP'P".  Draw  Q'P*  parallel  to  00', 
intersecting  the  theoretical  compression  curve  at  P'.  Q'P'  is  found 
by  measurement  to  be  0.9  of  A0f.  The  theoretical  terminal  pres- 
sure, 0'Q'f  measures  33  Ibs.,  the  specific  volume  at  that  pressure 

being  12.41  cubic  feet,  and  the  weight  per  cu.  ft.,  ^  ^  =  0.08051 

13,750  X  0.9  X  0.08051 

Ib.     Then.   -  -  =  23.5  Ibs.  steam  per  I.  H.  P. 

42.4 

per  hour. 

To  show  that  the  estimate  of  the  steam  consumption  per  I.  H.  P. 
per  hour  just  made  is  independent  of  the  size  and  speed  of  the 
engine,  we  will  assume  that  the  engine  from  which  the  diagram, 
Fig.  43,  was  taken  has  a  cylinder  20"  in  diameter,  a  stroke  of  2'  and 
was  making  200  revolutions  per  minute  when  the  diagram  was  taken. 


ECONOMY  OF  THE  STEAM  ENGINE  AND  BOILER          155 
Then,  3.1416  X  100  X  0.413  X  *X  200  X  2  X  60  X  0.194  g  ^  ^ 

of  steam  used  per  hour  when  measured  at  cut-off,  neglecting  the 

saving  by  compression. 

3.1416  X  100  X  0.3125  X  2  X  200  X  2  X  60  X  0.04307  _  1409  4  lbg 

144 
steam  saved  by  compression. 

Therefore,  8390  —  1409.4  =  6980.6  pounds  of  steam  used  per 
hour  when  measured  at  cut-off. 

T  TT  P  -  3-1416  X  10°  X  42-4  X  2  X  200  X  2  _  q99  Q0 
I.  H.  P.  -  33,000  m92' 

£*QO  A     /» 

Therefore,   owo   =21.617  pounds  of  steam  per  I.  H.  P.  per 


hour,  which  is  the  same  result  as  that  obtained  by  the  use  of  the 
constant  13,750. 

150.  Combining  Diagrams  of  Stage  Expansion  Engines.  —  Owing 
to  the  difference  in  the  initial  pressures  in  the  cylinders  of  stage  ex- 
p'ansion  engines,  springs  of  different  tensions  are  used  in  the  indi- 
cators, and  consequently  the  diagrams  of  the  several  cylinders  are 
to  different  scales. 

A  combination  on  the  scale  of  the  diagrams  of  a  stage  expan- 
sion engine  produces  a  single  diagram  which  exhibits  the  changes 
in  pressure  and  volume  the  steam  undergoes  from  the  moment  of 
its  admission  into  the  H.  P.  cylinder  until  its  final  release  from 
the  L.  P.  cylinder.  Such  a  diagram  enables  a  comparison  to  be 
made  between  the  work  actually  accomplished  and  that  theoretically 
due  to  the  initial  pressure  and  ratio  of  expansion,  under  the.  suppo- 
sition that  the  total  expansion  takes  place  in  the  L.  P.  cylinder. 
Owing  to  fundamental  defects  in  the  steam  engine  itself,  the  results 
obtained  from  the  combined  diagram  cannot  be  more  than  approxi- 
mate, but  the  process  of  the  combination  is  instructive,  and  if  care 
be  exercised  in  the  interpretation  of  the  results,  valuable  informa- 
tion concerning  the  general  performance  of  the  engine  may  be  ob- 
tained and  a  factor  determined  which  may  be  of  value  in  subsequent 
designs. 

In  the  process  of  combining  the  diagrams,  the  pressure  scale  of 
all  the  diagrams  is  the  same,  while  the  length  of  each  must  be 
such  as  to  represent  the  stroke-volume  of  the  cylinder  from  which  it 
was  taken.  It  is  important  that  the  clearance  volume  of  each  cylinj 


156  THE  ELEMENTS  OF  STEAM  ENGINEERING 

der,  expressed  in  percentages  of  stroke-volume,  be  added  to  the 
length,  of  each  diagram. 

The  diagrams,  Figs.  44,  45,  and  46,  are  from  a  triple-expansion 
engine,  the  cylinder  diameters  of  which  are  26.5",  39",  and  63",  and 
piston  stroke  26".  The  recorded  steam  pressures  were : 

At  H.  P.  cylinder 184.7  Ibs.  absolute. 

At  1st  receiver 93.7      "          " 

At  2d  receiver. .  .   29.7     " 


Diameter  of  piston  rods,  4.75";  vacuum,  25";  revolutions  per 
minute,  174.  Clearance:  H.  P.  Blinder,  17.28  per  cent;  I.  P.  cylin- 
der, 16.3  per  cent;  L.  P.  cylinder,  15  per  cent. 

^From  the  data  we  deduce  : 

Net  piston  areas  ____  542.28  sq.  in  ____  1216.56  sq.  in.  .  .  .3158.07 
sq.  in.     Ratios  of  net  piston  areas  referred  to  L.  P.  cylinder, 

1  :  5.819         1  :  2.596,         1.000. 
Volumetric  cylinder  ratios  referred  to  L.  P.  cylinder, 


.  3158.07  X  1.15  .  - 
:  542.68  X  1.1728'  5' 


.  3158.07  X  1.15  .  9 
'  1216.56  X  1.163'  2' 


ECONOMY  OF  THE  STEAM  ENGINE  AND  BOILER 


157 


The  first  step  in  the  process  of  combination  is  to  superpose  the 
diagrams  from  the  two  ends  of  each  cylinder,  and  thus  obtain  the 
mean  full  line  diagrams  shown  in  Eigs.  47,  48,  and  49.  Erect  per- 
pendiculars to  the  atmospheric  lines,  touching  the  diagrams  at  each 


R 

7** 

^- 

T- 

,  — 

—  i^> 



^i 

n 

l\ 
1  V 

I           V 

\ 

/ 

'i 
i 

^ 

—  i  — 

i 

**•*. 

\ 

1    >. 

1  / 

'  V, 

k 

i 
* 

}, 

£z^_ 

a 


end.  Lay  off,  to  the  pressure  scale  of  each  diagram,  the  perfect 
vacuum  lines,  14.7  Ibs.  below  the  respective  atmospheric  lines.  Lay 
off  ao,  equal  to  the  respective  cylinder  clearance  percentage  of  ab, 
and  erect  the  clearance  lines  oc.  Select  the  point  x,  as  near  after 


158 


THE  ELEMENTS  or  STEAM  ENGINEERING 


the  actual  point  of  cut-off  as  the  eye  can  detect,  and  where,  it  is  fair 
to  assume,  the  expansion  is  according  to  Boyle's  law.  Construct 
the  hyperbolic  expansion  curves  for  each  diagram,  from  which  a  fair 
idea  may  be  obtained  as  to  the  character  of  the  actual  expansion  in 
each  cylinder.  Select  the  point  y,  as  near  after  the  L.  P.  exhaust 
closure  as  the  eye  can  detect,  and  construct  the  hyperbolic  compres- 
sion curve.  Erect  ordinates  and  obtain  the  mean  pressure  from 
each  diagram  in  the  usual  manner. 

y 


Draw  the  coordinate  axes  OX  and  OF,  Fig.  50,  representing  the 
lines  of  volume  and  pressure  (from  perfect  vacuum),  respectively. 
Assume,  arbitrarily,  the  pressure  scale  to  be  20  Ibs.  to  the  inch  (that 
of  the  L.  P.  diagram),  and  that  OB,  equal  10  inches,  represents  the 
volume  of  the  L.  P.  cylinder  and  its  clearance.  (Fig.  50  is  drawn 
to  one-third  scale.) 

Then,  in  lineal  measurement,  10  inches  represents  1.15  times  the 

volume  of  the  L.  P.  cylinder,  therefore  tr-y^  =  8.7",  represents  the 
volume  of  the  L.  P.  cylinder.     Lay  off  OB  =  10",  and  AB  —  8.7". 


ECONOMY  OF  THE  STEAM  ENGINE  AND  BOILER          159 

Then  10  —  8.7  —  1.3"  =  OA,  represents  the  clearance  of  the  L.  P. 
cylinder. 

=  3.895"  =  PM,  represents  the  volume  of  the  I.  P.  cylin- 


q    OQK 

der  plus  its  clearance;   therefore,    '„  ^  =  3.35"  —  TM,  represents 

the  volume  of  the  I.  P.  cylinder,  and  3.895  —  3.35  =  0.545"  =  FT, 
represents  the  clearance  of  the  I.  P.  cylinder. 

5    *025  —  1.7524"  =  OK,  represents  the  volume  of  the  I.  P.  cyl- 


inder plus  its  clearance;  therefore    '  „„„  =  1.4942"  =  RK,  repre- 

sents the  volume  of  the  H.  P.  cylinder,  and  1.7524—1.4942 
=  0.2582"  =  QR,  represents  the  clearance  of  the  H.  P.  cylinder. 

Divide  AB,  TM,  and  RK  each  into  ten  equal  parts,  and  erect 
ordinates  midway  between  the  divisions.  Transfer  to  these  ordin- 
ates  the  pressures  obtained  from  the  ordinates  of  Figs.  47,  48,  and 
49,  and  through  the  points  thus  obtained  draw  in  the  full  line  dia- 
grams of  Fig.  50.  They  will  be  the  diagrams  of  Figs.  47,  48,  and 
49,  drawn  to  the  pressure  scale  of  20  Ibs.  to  the  inch,  and  of  lengths 
corresponding  to  the  relative  volumes  of  the  cylinders. 

Transfer  the  point  x  of  Fig.  47  to  x'  of  Fig.  50,  and  determine 
the  theoretical  point  of  cut-off,  z,  to  produce  the  same  terminal 
pressure  from  an  initial  pressure  of  184.7  Ibs.  as  that  obtained  from 
the  cut-off  at  x*  from  the  initial  pressure  of  147.5  Ibs.  Construct 
the  hyperbolic  expansion  curve  zCS.  The  theoretical  terminal  pres- 
sure, BS,  is,  by  measurement,  17.6  Ibs.  This  may  be  checked  by 

applying  to  the  curve  the  equation,  xy  =  «-,  of  the  rectangular 
hyperbola:  thus,  OB  X  BS  =  ^P-.,  whence  BS  =  f 


=  17.64  Ibs.  to  the  scale  of  20  Ibs.  to  the  inch.  The  actual  terminal 
pressure  is  12.8  Ibs.,  determined  by  drawing  the  hyperbolic  expan- 
sion curve  of  the  L.  P.  cylinder. 

A  comparison  of  the  area  included  between  the  hyperbola  and  the 
axes  with  that  of  the  full  line  diagrams,  illustrates  the  losses  due  to 
wire-drawing,  resistances  in  the  passages,  "  drop,"  and  clearance^ 
It  will  be  noticed  that  the  I.  P.  diagram  overlaps  the  H.  P.  diagram. 
This  may  very  well  occur  in  any  instance,  for  any  one  point  on  the 
forward  pressure  line  of  one  diagram  does  not  correspond  with  the 


COMB/MED 
of  a 

TRIPLE  EXP0/V3/0/V  MARINE  ENG/NE 

**f 


** 


/.  Diameters  of  Cy/i n t/ers  / 
Z.D/amete.r  of  P/ston  ffoc/s  ' 

*  -r  -T-    ' 

J£/Ve/  P/'ston  /?re0s  ;  S42. 68°*,  /2J6.56 °",  3158.07°''. 

4. Faf/os  of/Ye/P/sfon  flreas;     /.'S.8/8      ,  /: 2.596,       /   i    . 
£.C/ear&/?ces  in  percent,          /7£8 ,     /6.3    t     /^"     . 
6.Vo/vmefr/c  Cy//nc/er  fot/os; /.'S.706ZS,  /: 2.5669,      / 
7.  Stroke  of  Pistons >  2  6  " 

8.  Re.  volutions  per  minute  ,  /  7^ 

d^Steam  Pressure  at  Engine ,  I84-.  7  Ibs  absolute 

/O.      »  «          in  /st  Receiver,       33. 7    "  " 

//.      *  "          "  2nd     «        f        29.7    " 

JZ.  Terminal  Pressure  by  Hypenjolic  Cun/e,  /  7  64  * 

l3./?cfao/  Termina/  Pressure  ,  /%•  8     " 

/4.  Theoretic*?/  n  umber  of  Expansions,  IO.  4-  7 
of  Expansions,  //. 


M.  E.  P. 

1.  H.   P. 

On  Each 
P/'aton 

Reduced  fo 
i.PPis+on 

Each 
Cy/inc/er 

ffe/ati've 
per  cenf: 

/6. 

H.PCylinder 

73 

JZ.545 

90S 

Z9.888 

17, 

I.P.     •• 

36.3 

13.983 

IO/9 

33.652 

18. 

L.P     •> 

15.3 

I&.300 

IIO4 

36.460 

19. 

TOTAL 

41.828 

3028 

/0O.  OOO 

.  Tota/  /tf.F.f?  ob  fa/net/  from 


.  Mean  Pressure  Factor 


-     0.  7O3  6 


23. 


ECONOMY  or  THE  STEAM  ENGINE  AND  BOILER          161 

point  of  the  back  pressure  line  of  the  diagram  of  the  preceding 
cylinder  found  on  the  same  ordinate.  The  point  of  correspondence 
is  a  matter  of  calculation,  depending  on  the  relative  piston  positions. 

Theoretical  number  of  expansions  =  jy^  —  10.47. 

147  5 
Actual  number  of  expansions  —   -^  g  =  11.52. 

The  equivalent  of  the  mean  effective  pressures  on  the  pistons  of  the 
high  and  intermediate-pressure  cylinders,  when  referred  to  the 

73  36  3 

L.    P.    piston,    are,   respectively,   r-T     =  12.545    Ibs.    and 


=  13.983  Ibs.  The  total  mean  effective  pressure,  referred  to  the 
L.  P.  piston,  is  then  12.545  +  13.983  +  15.300  =  41.828  Ibs.  The 
theoretical  mean  effective  pressure,  due  to  an  initial  pressure  of 
184.7  Ibs.  and  a  ratio  of  expansion  of  10.47,  is  : 

184.7(1  +  log.  10.47)  _  184.7  X  3.37  _.  59  lg  » 
10.47  10.47 

41  828 
Mean  pressure  factor  =      '  .„  =  0.7036. 

The  indicated  horse-powers  developed  in  the  high,  intermediate, 
and  low-pressure  cylinders  are  905,  1019,  and  1104,  respectively, 
making  a  total  for  the  engine  of  3028  I.  H.  P. 

The  weight  of  a  cubic  foot  of  steam  at  pressure  of  17.64  Ibs.  is 
found  from  the  table  to  be  0.0446  Ib.  Then,  the  approximate  steam 
consumption  per  I.  H.  P.  per  hour  is  : 

13,750  X  0.0446  X  0.925  _    „       „ 
41.8*8 

151.  The  Dynamometer.  —  In  order  to  find  the  brake  horse-power 
of  an  engine,  or  the  useful  power  delivered  to  the  shaft,  independent 
of  the  power  absorbed  in  friction  in  driving  the  engine  itself,  some 
form  of  dynamometer  is  used.  Fig.  51  is  a  representation  of  an 
absorption  dynamometer,  known  as  the  Prony  Brake.  A  drum  pul- 
ley, A,  keyed  on  the  shaft,  (7,  has  a  strap,  88,  partially  encircling 
it.  Bolted  to  the  strap,  and  bearing  on  the  pulley,  is  a  series  of 
blocks,  &&,  which  have  spaces  intervening  between  them.  A  wooden 
beam,  B,  has  a  wooden  shoe,  D,  which  also  bears  on  the  pulley,  A. 
The  weight,  w,  is  adjustable,  and  serves  to  balance  the  preponder- 
ance of  that  part  of  the  beam  to  the  right  of  the  center  of  the  shaft. 
11 


162  THE  ELEMENTS  OF  STEAM  ENGINEERING 

If  the  strap,  SS,  be  tightened  about  the  pulley,  A,  by  means  of 
the  nuts,  nn>  one  of  two  things  must  happen  when  the  engine  starts 
and  the  shaft  revolves  in  the  direction  of  the  arrow ;  either  the  grip 
of  the  strap  must  be  sufficient  to  carry  the  blocks  and  beam  round 
with  the  pulley  or  the  pulley  must  slip  round  over  the  blocks.  In 
practice  the  nuts  are  so  adjusted  that  the  pulley  just  slips  round  and 
the  beam  kept  horizontal.  The  resistance  to  the  carrying  of  the 
beam  and  blocks  round  with  the  pulley  is  the  pressure,  W,  made 
to  act  on  a  platform  scales  as  shown,  so  that  its  amount  may  be 
measured.  The  resistance  to  the  pulley  slipping  round  in  the  strap 
is  the  friction,  F. 

At  the  moment  of  slipping  we  shall  have,  taking  moments  about 

the  center  of  the  shaft,  rF  =  aW,  whence  F  = ,  in  which  r  is 


the  radius  of  the  pulley  and  a  the  distance  from  the  center  of  the 
shaft  to  the  line  of  pressure  on  the  scales. 

For  one  revolution  of  the  shaft,   the  work   done   is   F  X  2?rr 

a,  W 
=  -  X  Zirr  =  %iraW  foot-pounds,  when  a  is  measured  in  feet 

and  W  in  pounds.     For  n  revolutions  the  work  is  27rnaW,  and  the 
horse-power  is 


In  practice  the  pulley  may  be  kept  cool  by  circulating  water 
within  it.  If  both  the  rubbing  surfaces  are  metallic,  they  should 
be  freely  lubricated.  An  iron  pulley  rubbing  over  wooden  blocks, 
as  in  Fig.  51,  requires  little  lubrication. 

152.  Engine  and  Boiler  Tests.  —  The  primary  object  in  testing  a 
steam  engine  is  to  determine  the  cost  of  the  power.  The  mean 
effective  pressure  as  determined  from  the  indicator  diagram  is  the 
exponent  of  the  power,  and  the  cost  is  expressed  in  thermal  units, 
in  pounds  of  steam,  or  in  pounds  of  coal  per  unit  of  power. 


ECONOMY  OF  THE  STEAM  ENGINE  AND  BOILER          163 

If  the  measure  is  to  be  in  pounds  of  steam  per  I.  H.  P.  per  hour, 
there  are  two  methods — one  for  the  surface-condensing  and  one  for 
the  non-condensing  and  for  the  jet-condensing  types  of  engines. 
With  the  surface-condensing  engine  the  water  resulting  from  the 
condensation  of  the  exhaust  steam  is  carefully  weighed,  allowance 
being  made  for  all  steam  used  for  other  purposes  incidental  to  the 
operation  of  the  engine,  and  correction  being  made  for  moisture. 
An  hour's  duration  of  such  a  test  will  give  accurate  results.  In 
testing  non-condensing  and  jet-condensing  engines  the  weight  of 
steam  per  I.  H.  P.  per  hour  is  determined  by  weighing  the  feed 
water  supplied  to  the  boilers,  always  making  allowance  for  the 
steam  used  for  purposes  other  than  for  driving  the  engine,  such,  for 
example,  as  for  pumping,  for  jackets,  for  heating,  or  for  calori- 
meter tests. 

When  the  measure  is  made  in  pounds  of  coal  per  I.  H.  P.  per  hour 
the  test  must  be  one  combined  of  the  engine  and  the  boilers,  in 
which  all  the  precautions  stipulated  for  a  boiler  test  are  to  be 
observed,  and  the  coal  carefully  weighed.  The  duration  of  such 
trials  is  from  12  to  24  hours. 

Tests  may  be  made  to  determine  the  economy  derived  from  the 
use  of  special  appurtenances,  such  as  jackets  and  economizers. 
The  use  of  feed  water  heaters  with  non-condensing  engines  may 
effect  a  saving  in  fuel  as  great  as  12  per  cent. 

153.  The  installation  of  engine  and  boiler  plants  is  usually  done 
by  contract,  and  the  only  means  of  determining  whether  the  stipu- 
lations of  the  contract  have  been  fulfilled  is  to  test  the  engine  and 
boilers  as  a  system.     Preceding  the  commencement  of  such  a  test 
the  engine  should  be  run  for  several  hours  under  the  test  condi- 
tions, and  it  is,  of  course,  very  necessary  that  the  conditions  remain 
constant  during  the  test.     This  is  particularly  true  with  respect  to 
the  steam  pressure.     It  is  equally  important  that  the  measuring 
instruments,  such  as  thermometers,  gauges,  indicators,  and  scales 
should  be  standardized  before  the  commencement  of  a  test. 

Everything  pertaining  to  the  cost  of  the  operation  and  mainten- 
ance of  a  plant  enters  into  its  economy,  and  it  is  the  province  of  the 
engineer  to  obtain  the  best  results  at  the  minimum  of  cost. 

154.  The  primary  object  in  testing  a  boiler  is  to  determine  the 
number  of  pounds  of  water  evaporated  per  pound  of  coal,  and  the 
conditions  of  the  test  should  be  such  that  the  boiler  be  not  driven 


164  THE  ELEMENTS  OF  STEAM  ENGINEERING 

above  what  is  supposed  to  be  its  rated  capacity.  To  drive  a  boiler 
to  its  utmost  would  be  a  test  of  its  capacity,  but  since  it  is  far 
from  economical  to  so  work  a  boiler,  such  tests  are  not  of  common 
occurrence. 

At  the  meeting  in  1899  of  the  American  Society  of  Mechanical 
Engineers,  a  revised  code  for  conducting  steam  boiler  trials  was 
adopted  and  termed  the  Code  of  1899. 

The  Short  Code  itself,  together  with  some  extracts  from  the 
report  of  the  committee  as  published  in  the  Journal  of  the  Ameri- 
can Society  of  Naval  Engineers,  will  here  be  given. 

EULES  FOR  CONDUCTING  BOILER  TRIALS — CODE  OF  1899. 

1.  Determine  at  the  outset  the  specific  object  of  the  proposed 
trial,  whether  it  be  to  ascertain  the  capacity  of  the  boiler,  its  effi- 
ciency as  a  steam  generator,  its  efficiency  and  its  defects  under 
usual  working  conditions,  the  economy  of  some  particular  kind  of 
fuel,  or  the  effect  of  changes  of  design,  proportion,  or  operation; 
and  prepare  for  the  trial  accordingly. 

2.  Examine  the  boiler,  both  outside  and  inside;    ascertain  the 
dimensions  of  grates,  heating  surfaces,  and  all  important  parts; 
and  make  a  full  record,  describing  the  same,  and  illustrating  special 
features  by  sketches.     The  area  of  heating  surface  is  to  be  com- 
puted from  the  surfaces  of  shells,  tubes,  furnaces,  and  fire  boxes 
in  contact  with  the  fire  or  hot  gases.     The  outside  diameter  of 
water  tubes  and  the  inside  diameter  of  fire  tubes  are  to  be  used  in 
the  computation.     All  surfaces  below  the  mean  water  level  which 
have  water  on  one  side  and  products  of  combustion  on  the  other  are 
to  be  considered  as  water-heating  surface,  and  all  surfaces  above 
the  mean  water  level  which  have  steam  on  one  side  and  products 
of  combustion  on  the  other  are  to  be  considered  as  superheating 
surface. 

3.  Notice  the  general  condition  of  the  boiler  and  its  equipment, 
and  record  such  facts  in  relation  thereto  as  bear  upon  the  objects 
in  view. 

If  the  object  of  the  trial  is  to  ascertain  the  maximum  economy 
or  capacity  of  the  boiler  as  a  steam  generator,  the  boiler  and  all  its 
appurtenances  should  be  put  in  first-class  condition.  Clean  the 
heating  surface  inside  and  outside,  remove  clinkers  from  the  grates 
and  from  the  sides  of  the  furnace.  Eemove  all  dust,  soot,  and 
ashes  from  the  chambers,  smoke  connections,  and  flues.  Close  air 


ECONOMY  OF  THE  STEAM  ENGINE  AND  BOILER          165 

leaks  in  the  masonry  and  poorly  fitted  cleaning  doors.  See  that  the 
damper  will  open  wide  and  close  tight.  Test  for  air  leaks  by  firing 
a  few  shovelfuls  of  smoky  fuel  and  immediately  closing  the  damper, 
observing  the  escape  of  smoke  through  the  crevices,  or  by  passing 
the  flame  of  a  candle  over  cracks  in  the  brickwork. 

4.  Determine  the  character  of  the  coal  to  be  used.     For  tests  of 
the  efficiency  or  capacity  of  the  boiler  for  comparison  with  other 
boilers  the  coal  should,  if  possible,  be  of  some  kind  which  is  com- 
mercially regarded  as  a  standard.     For  New  England  and  that 
portion  of  the  country  east  of  the  Allegheny  Mountains,  a  good 
anthracite  egg  coal,  containing  not  over  10  per  cent  of  ash,  and 
semi-bituminous  Clearfield   (Pa.),  Cumberland   (Md.),  and  Poca- 
hontas    (Va.)    coals  are  thus  regarded.     West  of  the   Allegheny 
Mountains,  Pocahontas  (Va.)  and  New  River  (W.  Va.)  semi-bitu- 
minous,   and   Youghiogheny   or    Pittsburg   bituminous    coals    are 
recognized  as  standards.     (These  coals  are  selected  because  they 
are  about  the  only  coals  which  possess  the  essentials  of  excellence 
of  quality,  adaptability  to  various  kinds  of  furnaces,  grates,  boilers, 
and  methods  of  firing,  and  wide  distribution  and  general  accessi- 
bility in  the  markets.)     There  is  no  special  grade  of  coal  mined  in 
the  Western  States  which  is  widely  recognized  as  of  superior  quality 
or  considered  as  a  standard  coal  for  boiler  testing.     Big  Muddy 
lump,  an  Illinois  coal  mined  in  Jackson  County,  111.,  is  suggested 
as  being  of  sufficiently  high  grade  to  answer  these  requirements  in 
districts  where  it  is  more  conveniently  obtainable  than  the  other 
coals  mentioned  above. 

For  tests  made  to  determine  the  performance  of  a  boiler  with 
a  particular  kind  of  coal,  such  as  may  be  specified  in  a  contract  for 
the  sale  of  a  boiler,  the  coal  used  should  not  be  higher  in  ash  and 
in  moisture  than  that  specified,  since  increase  in  ash  and  moisture 
above  a  stated  amount  is  apt  to  cause  a  falling  off  of  both  capacity 
and  economy  in  greater  proportion  than  the  proportion  of  such 
increase. 

5.  Establish  the  correctness  of  all  apparatus  used  in  the  test  for 
weighing  and  measuring.     These  are : 

(a)   Scales  for  weighing  coal,  ashes,  and  water. 

(&)  Tanks  or  water  meters  for  measuring  water.  Water  meters, 
as  a  rule,  should  be  used  only  as  a  check  on  other  measurements. 
For  accurate  work  the  water  should  be  weighed  or  measured  in  a 
tank. 


166  THE  ELEMENTS  OF  STEAM  ENGINEERING 

(c)  Thermometers  and  pyrometers  for  taking  temperatures  of 
air,  steam,  feed  water,  waste  gases,  etc. 

(d)  Pressure  gauges,  draft  gauges,  etc. 

The  kind  and  location  of  the  various  pieces  of  testing  apparatus 
must  be  left  to  the  judgment  of  the  person  conducting  the  test; 
always  keeping  in  mind  the  main  object,  i.e.,  to  obtain  authentic 
data. 

6.  See  that  the  boiler  is  thoroughly  heated  before  the  trial,  to 
its  usual  working  temperature.     If  the  boiler  is  new  and  of  a  form 
provided  with  a  brick  setting,  it  should  be  in  regular  use  at  least  a 
week  before  the  trial,  so  as  to  dry  and  heat  the  walls.     If  it  has 
been  laid  off  and  become  cold,  it  should  be  worked  before  the  trial 
until  the  walls  are  well  heated. 

7.  The  boiler  and  connections  should  be  proved  to  be  free  from 
leaks  before  beginning  a  test,  and  all  water  connections,  including 
blow  and  extra  feed  pipes,  should  be  disconnected,  stopped  with 
blank  flanges,  or  bled  through  special  openings  beyond  the  valves, 
except  the  particular  pipe  through  which  water  is  to  be  fed  to  the 
boiler  during  the  trial.     During  the  test  the  blow-off  and  feed 
pipes  should  remain  exposed  to  view. 

If  an  injector  is  used,  it  should  receive  steam  directly  through  a 
felted  pipe  from  the  boiler  being  tested. 

If  the  water  is  metered  after  it  passes  the  injector,  its  tempera- 
ture should  be  taken  at  the  point  where  it  leaves  the  injector.  If 
the  quantity  is  determined  before  it  goes  to  the  injector,  the  tem- 
perature should  be  determined  on  the  suction  side  of  the  injector, 
and  if  no  change  of  temperature  occurs  other  than  that  due  to  the 
injector,  the  temperature  thus  determined  is  properly  that  of  the 
feed  water.  When  the  temperature  changes  between  the  injector 
and  the  boiler,  as  by  the  use  of  a  heater  or  by  radiation,  the  tem- 
perature at  which  the  water  enters  and  leaves  the  injector  and  that 
at  which  it  enters  the  boiler  should  all  be  taken.  In  that  case  the 
weight  to  be  used  is  that  of  the  water  leaving  the  injector,  com- 
puted from  the  heat  units  if  not  directly  measured,  and  the  tem- 
perature, that  of  the  water  entering  the  boiler. 

Let  w  =  weight  of  water  entering  the  injector. 
x  =.  weight  of  steam  entering  the  injector. 
jk±  =  heat  units  per  pound  of  water  entering  the  injector. 
7i2  =  heat  units  per  pound  of  steam  entering  the  injector. 
h3  =  heat  units  per  pound  of  water  leaving  the  injector. 


ECONOMY  OF  THE  STEAM  ENGINE  AND  BOILER          16 

Then,  w(h3  —  h±)  =  units  gained  by  the  water. 
x(h2  —  h3)  —  units  lost  by  the  steam. 

Therefore,  'x  = 

See  that  the  steam  main  is  so  arranged  that  water  of  condensa- 
tion cannot  run  back  into  the  boiler. 

8.  Duration  of  the  Test. — For  tests  made  to  ascertain  either  the 
maximum  economy  or  the  maximum  capacity  of  a  boiler,  irrespec- 
tive of  the  particular  class  of  service  for  which  it  is  regularly  used, 
the  duration  should  be  at  least  10  hours  of  continuous  running. 
If  the  rate  of  combustion  exceeds  25  pounds  of  coal  per  square  foot 
of  grate  surface  per  hour,  it  may  be  stopped  when  a  total  of  300 
pounds  of  coal  have  been  burned  per  square  foot  of  grate. 

In  cases  where  the  service  requires  continuous  running  for  the 
whole  24  hours  of  the  day,  with  shifts  of  firemen  a  number  of 
times  during  that  period,  it  is  well  to  continue  the  test  for  at  least 
24  hours. 

When  it  is  desired  to  ascertain  the  performance  under  the  work- 
ing conditions  of  practical  running,  whether  the  boiler  be  regularly 
in  use  24  hours  a  day  or  only  a  certain  number  of  hours  out  of 
each  24,  the  fires  being  banked  the  balance  of  the  time,  the  duration 
should  not  be  less  than  24  hours. 

9.  Starting  and  Stopping  a  Test. — The  conditions  of  the  boiler 
and  furnace  in  all  respects  should  be,  as  nearly  as  possible,  the 
same  at  the  end  as  at  the  beginning  of  the  test.     The  steam  pres- 
sure should  be  the  same;  the  water  level  the  same;  the  fire  upon 
the  grates  should  be  the  same  in  quantity  and  condition;  and  the 
walls,  flues,  etc.,  should  be  of  the  same  temperature.     Two  methods 
of  obtaining  the  desired  equality  of  conditions  of  the  fire  may  be 
used,  viz :  those  which  were  called  in  the  Code  of  1885  "  the  stand- 
ard method  "  and  "  the  alternate  method,"  the  latter  being  employed 
where  it  is  inconvenient  to  make  use  of  the  standard  method. 

10.  Standard  Method  of  Starting  and  Stopping  a  Test. — Steam 
being  raised  to  the  working  pressure,  remove  rapidly  all  the  fire 
from  the  grate,  close  the  damper,  clean  the  ash  pit,  and  as  quickly 
as  possible  start  a  new  fire  with  weighed  wood  and  coal,  noting  the 
time  and  the  water  level  while  the  water  is  in  a  quiescent  state,  just 
before  lighting  the  fire. 

At  the  end  of  the  test  remove  the  whole  fire,  which  has  burned 
low,  clean  the  grates  and  ash  pit,  and  note  the  water  level  when  the 


168  THE  ELEMENTS  OF  STEAM  ENGINEERING 

water  is  in  a  quiescent  state,  and  record  the  time  of  hauling  the  fire. 
The  water  level  should  be  as  nearly  as  possible  the  same  as  at  the 
beginning  of  the  test.  If  it  is  not  the  same,  a  correction  should  be 
made  by  computation,  and  not  by  operating  the  pump  after  the 
test  is  completed. 

11.  Alternate  Method  of  Starting  and  Stopping  a  Test. — The 
boiler  being  thoroughly  heated  by  a  preliminary  run,  the  fires  are 
to  be  burned  low  and  well  cleaned.     Note  the  amount  of  coal  left 
on  the  grate  as  nearly  as  it  can  be  estimated ;  note  the  pressure  of 
steam  and  the  water  level:     Note  the  time,  and  record  it  as  the 
starting  time.     Fresh  coal  which  has  been  weighed  should  now  be 
fired.     The  ash  pits  should  be  thoroughly  cleaned  at  once  after 
starting.     Before  the  end  of  a  test  the  fires  should  be  burned  low, 
just  as  before  the  start,  and  the  fires  cleaned  in  such  manner  as  to 
leave  a  bed  of  coal  on  the  grates  of  the  same  depth,  and  in  the  same 
condition  as  at  the  start.     When  this  stage  is  reached,  note  the 
time  and  record  it  as  the  stopping  time.     The  water  level  and 
steam  pressures  should  previously  be  brought  as  nearly  as  possible 
to  the  same  point  as  at  the  start.     If  the  water  level  is  not  the  same 
as  at  the  start,  a  correction  should  be  made  by  computation,  and 
not  by  operating  the  pump  after  the  test  is  completed. 

12.  Uniformity  of  Conditions. — In  all  trials  made  to  ascertain 
maximum  economy  or  capacity,  the  conditions  should  be  main- 
tained uniformly  constant.     Arrangements  should  be  made  to  dis- 
pose of  the  steam  so  that  the  rate  of  evaporation  may  be  kept  the 
same  from  beginning  to  end.     This  may  be  accomplished  in  a 
single  boiler  by  carrying  the  steam  through  a  waste  steam  pipe,  the 
discharge  from  which  can  be  regulated  as  desired.     In  a  battery  of 
boilers,  in  which  only  one  is  tested,  the  draft  may  be  regulated  on 
the  remaining  boilers,  leaving  the  test  boiler  to  work  under  a  con- 
stant rate  of  production. 

Uniformity  of  conditions  should  prevail  as  to  the  pressure  of 
steam,  the  height  of  water,  the  rate  of  evaporation,  the  thickness 
of  fire,  the  times  of  firing  and  quantity  of  coal  fired  at  one  time, 
and  as  to  the  intervals  between  the  times  of  cleaning  the  fires. 

The  method  of  firing  to  be  carried  on  in  such  tests  should  be 
dictated  by  the  expert  or  person  in  responsible  charge  of  the  test, 
and  the  method  adopted  should  be  adhered  to  by  the  fireman 
throughout  the  test. 


ECONOMY  OF  THE  STEAM  ENGINE  AND  BOILER          169 

13.  Keeping  the  Records. — Take  note  of  every  event  connected 
with  the  progress  of  the  trial,  however  unimportant  it  may  appear. 
Eecord  the  time  of  every  occurrence  and  the  time  of  taking  every 
weight  and  every  observation. 

The  coal  should  be  weighed  and  delivered  to  the  fireman  in  equal 
proportions,  each  sufficient  for  not  more  than  one  hour's  run,  and 
a  fresh  portion  should  not  be  delivered  until  the  previous  one  has 
all  been  fired.  The  time  required  to  consume  each  portion  should 
be  noted,  the  time  being  recorded  at  the  instant  of  firing  the  last 
of  each  portion.  It  is  desirable  that  at  the  same  time  the  amount 
of  water  fed  into  the  boiler  should  be  accurately  recorded,  includ- 
ing the  height  of  the  water  in  the  boiler,  and  the  average  pressure 
of  steam  and  temperature  of  feed  during  the  time.  By  thus  re- 
cording the  amount  of  water  evaporated  by  successive  portions  of 
coal,  the  test  may  be  divided  into  several  periods  if  desired,  and 
the  degree  of  uniformity  of  combustion,  evaporation,  and  economy 
analyzed  for  each  period.  In  addition  to  these  records  of  the  coal 
and  the  feed  water,  half-hourly  observations  should  be  made  of 
the  temperature  of  the  feed  water,  of  the  flue  gases,  of  the  external 
air  in  the  boiler  room,  of  the  temperature  of  the  furnace  when  a 
furnace  pyrometer  is  used,  also  of  the  pressure  of  steam  and  of  the 
readings  of  the  instruments  for  determining  the  moisture  in  the 
steam.  A  log  should  be  kept  on  properly  prepared  blanks  containing 
columns  for  record  of  the  various  observations. 

When  the  "  standard  method  "  of  starting  and  stopping  the  test 
is  used,  the  hourly  rate  of  combustion  and  of  evaporation  and  the 
horse-power  should  be  computed  from  the  records  taken  during  the 
time  when  the  fires  are  in  active  condition.  This  time  is  some- 
what less  than  the  actual  time  which  elapses  between  the  beginning 
and  end  of  the  run.  The  loss  of  time  due  to  kindling  the  fire  at 
the  beginning  and  burning  it  out  at  the  end  makes  this  course 
necessary. 

14.  Quality  of  Steam. — The  percentage  of  moisture  in  the  steam 
should  be  determined  by  the  use  of  either  a  throttling  or  a  separat- 
ing steam  calorimeter.     The  sampling  nozzle  should  be  placed  in 
the  vertical  steam  pipe  rising  from  the  boiler.     It  should  be  made 
of  -J-inch  pipe,  and  should  extend  across  the  diameter  of  the  steam 
pipe  to  within  half  an  inch  of  the  opposite  side,  being  closed  at  the 
end  and  perforated  with  not  less  than  twenty  -J-inch  holes  equally 
distributed  along  and  around  its  cylindrical  surface,  but  none  of 


170  THE  ELEMENTS  OF  STEAM  ENGINEERING 

these  holes  should  be  nearer  than  J  inch  to  the  inner  side  of  the 
steam  pipe.  The  calorimeter  and  the  pipe  leading  to  it  should  be 
well  covered  with  felting.  Whenever  the  indications  of  the  throt- 
tling or  separating  calorimeter  show  that  the  percentage  of  moisture 
is  irregular,  or  occasionally  in  excess  of  3  per  cent,  the  results 
should  be  checked  by  a  steam  separator  placed  in  the  steam  pipe  as 
close  to  the  boiler  as  convenient,  with  a  calorimeter  in  the  steam 
pipe  just  beyond  the  outlet  from  the  separator.  The  drip  from  the 
separator  should  be  caught  and  weighed,  and  the  percentage  of 
moisture  computed  therefrom  added  to  that  shown  by  the  calori- 
meter. 

Superheating  should  be  determined  by  means  of  a  thermometer 
placed  in  a  mercury-well  inserted  in  the  steam  pipe.  The  degree  of 
superheating  should  be  taken  as  the  difference  between  the  reading 
of  the  thermometer  for  superheated  steam  and  the  readings  of  the 
same  thermometer  for  saturated  steam  at  the  same  pressure  as  deter- 
mined by  a  special  experiment,  and  not  by  reference  to  steam  tables. 

15.  Sampling  the  Coal  and  Determining  its  Moisture. — As  each 
barrow  load  or  fresh  portion  of  coal  is  taken  from  the  coal  pile,  a 
representative  shovelful  is  selected  from  it  and  placed  in  a  barrel 
or  box  in  a  cool  place  and  kept  until  the  end  of  the  trial.  The 
samples  are  then  mixed  and  broken  into  pieces  not  exceeding  one 
inch  in  diameter,  and  reduced  by  the  process  of  repeated  quarter- 
ings  and  crushings  until  a  final  sample  weighing  about  five  pounds  is 
obtained,  and  the  size  of  the  larger  pieces  are  such  that  they  will 
pass  through  a  sieve  with  J-inch  meshes.  From  this  sample  two 
one-quart,  airtight,  glass  preserving  jars,  or  other  airtight  vessels 
which  will  prevent  the  escape  of  moisture  from  the  sample,  are 
to  be  promptly  filled,  and  these  samples  are  to  be  kept  for  subse- 
quent determination  of  moisture  and  of  heating  value  and  for 
chemical  analyses.  During  the  process  of  quartering,  when  the 
sample  has  been  reduced  to  about  100  pounds,  a  quarter  to  a  half 
of  it  may  be  taken  for  an  approximate  determination  of  moisture. 
This  may  be  made  by  placing  it  in  a  shallow  iron  pan,  not  over  three 
inches  deep,  carefully  weighing  it,  and  setting  the  pan  in  the 
hottest  place  than  can  be  found  on  the  brickwork  of  the  boiler 
setting  or  flues,  keeping  it  there  for  at  least  twelve  hours,  and  then 
weighing  it.  The  determination  of  moisture  thus  made  is  believed 
to  be  approximately  accurate  for  anthracite  and  semi-bituminous 
coals,  and  also  for  Pittsburg  or  Youghiogheny  coal;  but  it  cannot 


ECONOMY  OF  THE  STEAM  ENGINE  AND  BOILER          171 

be  relied  upon  for  coals  mined  west  of  Pittsburg,  or  for  other  coals 
containing  inherent  moisture.  For  these  latter  coals  it  is  im- 
portant that  a  more  accurate  method  be  adopted.  The  method 
recommended  by  the  committee  for  all  accurate  tests,  whatever  the 
character  of  the  coal,  is  described  as  follows: 

Take  one  of  the  samples  contained  in  the  glass  jars,  and  subject 
it  to  a  thorough  air-drying,  by  spreading  it  in  a  thin  layer  and 
exposing  it  for  several  hours  to  the  atmosphere  of  a  warm  room, 
weighing  it  before  and  after,  thereby  determining  the  quantity  of 
surface  moisture  it  contains.  Then  crush  the  whole  of  it  by  run- 
ning it  through  an  ordinary  coffee  mill  adjusted  so  as  to  produce 
somewhat  coarse  grains  (less  than  ^  inch),  thoroughly  mix  the 
crushed  sample,  select  from  it  a  portion  of  from  10  to  50  grams, 
weigh  it  in  a  balance  which  will  easily  show  a  variation  as  small 
as  1  part  in  1000,  and  dry  it  in  an  air  or  sand  bath  at  a  temperature 
between  240°  and  280°  Fahrenheit  for  one  hour.  Weigh  it  and 
record  the  loss,  then  heat  and  weigh  it  again  repeatedly,  at  inter- 
vals of  an  hour  or  less,  until  the  minimum  weight  has  been  reached 
and  the  weight  begins  to  increase  by  oxidation  of  a  portion  of  the 
coal.  The  difference  between  the  original  and  the  minimum  weight 
is  taken  as  the  moisture  in  the  air-dried  coal.  This  moisture  test 
should  preferably  be  made  on  duplicate  samples,  and  the  results 
should  agree  within  0.3  to  0.4  of  1  per  cent.,  the  mean  of  the  two 
determinations  being  taken  as  the  correct  result.  The  sum  of  the 
percentage  of  moisture  thus  found  and  the  percentage  of  surface 
moisture  previously  determined  is  the  total  moisture. 

16.  Treatment  of  Ashes  and  Refuse. — The  ashes  and  refuse  are 
to  be  weighed  in  a  dry  state.     If  it  is  found  desirable  to  show  the 
principal  characteristics  of  the  ash,  a  sample  should  be  subjected 
to  a  proximate  analysis  and  the  actual  amount  of  incombustible 
material  determined.     For  elaborate  trials  a  complete  analysis  of 
the  ash  and  refuse  should  be  made. 

17.  Calorific  Tests  and  Analysis  of  Coal. — The  quality  of  the 
fuel  should  be  determined  either  by  heat  test  or  by  analysis,  or  by 
both. 

The  rational  method  of  determining  the  total  heat  of  combustion 
is  to  burn  the  sample  of  coal  in  an  atmosphere  of  oxygen  gas,  the 
coal  to  be  sampled  as  directed  in  Article  15  of  this  code. 

The  chemical  analysis  of  the  coal  should  be  made  only  by  an 
expert  chemist.  The  total  heat  of  combustion  computed  from  the 


172  THE  ELEMENTS  OF  STEAM  ENGINEERING 

results  of  the  ultimate  analysis  may  be  obtained  by  the  use  of  Du- 
long^s  formula  (with  constants  modified  by  recent  determinations), 
viz. :  14,600  C  +  62,000  (H  —  §  )  +  4000  S,  in  which  (7,  H,  0, 
and  8  refer  to  the  proportions  of  carbon,  hydrogen,  oxygen,  and 
sulphur,  respectively,  as  determined  by  the  ultimate  analysis. 

It  is  desirable  that  a  proximate  analysis  should  be  made,  thereby 
determining  the  relative  proportions  of  volatile  matter  and  fixed 
carbon.  These  proportions  furnish  an  indication  of  the  leading 
characteristics  of  the  fuel,  and  serve  to  fix  the  class  to  which  it 
belongs.  As  an  additional  indication  of  the  characteristics  of  the 
fuel,  the  specific  gravity  should  be  determined. 

(Articles  18,  19,  and  20  are  omitted,  as  they  relate  to  "  Analysis 
of  Flue  Gases,"  "Smoke  Observations,"  and  "Miscellaneous," 
which  are  deemed  unnecessary  in  the  use  of  the  short  form  of  the 
code.) 

21.  Calculations  for  Efficiency. — Two  methods  of  defining  and 
calculating  the  efficiency  of  a  boiler  are  recommended.  They  are: 

Heat  absorbed  per  Ib.  combustible 

1.  Efficiency  of  the  boiler  =  -^-^ — ^ 1 j-.ii       — t — TvT' 

Calorific  value  of  1  Ib.  combustible 

Heat  absorbed  per  Ib.  coal 

2.  Efficiency  of  boiler  and  grate  =.  Calorific  value  of  1  Ib.  coal' 

The  first  of  these  is  sometimes  called  the  efficiency  based  on  com- 
bustible, and  the  second  the  efficiency  based  on  coal.  The  first  is 
recommended  as  a  standard  of  comparison  for  all  tests,  and  this  is 
the  one  which  is  understood  to  be  referred  to  when  the  word  "  effi- 
ciency "  alone  is  used  without  qualification.  The  second,  however, 
should  be  included  in  a  report  of  a  test,  together  with  the  first, 
whenever  the  object  of  the  test  is  to  determine  the  efficiency  of  the 
boiler  and  furnace  together  with  the  grate  (or  mechanical  stoker), 
or  to  compare  different  furnaces,  grates,  fuels,  or  methods  of  firing. 

The  heat  absorbed  per  pound  of  combustible  (or  per  pound  of 
coal)  is  to  be  calculated  by  multiplying  the  equivalent  evaporation 
from  and  at  212°  per  pound  of  combustible  (or  coal)  by  965.7. 

DATA  AND  KESULTS  OF  EVAPORATIVE  TEST. 

Arranged  in  accordance  with  the  Short  Form  advised  by  the 
Boiler  Test  Committee  of  the  American  Society  of  Mechanical  En- 
gineers, Code  of  1899. 


ECONOMY  OF  THE  STEAM  ENGINE  AND  BOILER          173 


Made  by on boiler,  at 

to  determine 

Kind  of  fuel 

Kind  of  furnace 

Method  of  starting  the  test,  "  standard  "  or  "  alternate  " 

Grate  surface sq.  f t. 

Water  heating  surface " 

Superheating  surface " 

Total  Quantities. 

1.  Date  of  trial 

2.  Duration  of  trial hours. 

3.  Weight  of  coal  as  fired,  including  equivalent  of 

wood  used  in  lighting  the  fire,  not  including  un- 
burnt  coal  withdrawn  from  furnace  at  times  of 
cleaning  and  at  end  of  test.  One  pound  of  wood 
is  taken  to  be  equal  to  0.4  pound  of  coal,  or,  in 
case  greater  accuracy  is  desired,  as  having  a 
heat  value  equivalent  to  the  evaporation  of  6 
pounds  of  water  from  and  at  212°  per  pound. 
(6  X  965.7  =  5794  B.  T.  U.)  The  term  «  as 
fired  "  means  in  its  actual  condition,  including 
moisture Ibs. 

4.  Percentage  of  moisture  in  coal per  cent. 

5.  Total  weight  of  dry  coal  consumed Ibs. 

6.  Total  ash  and  refuse " 

7.  Percentage  of  ash  and  refuse  in  dry  coal per  cent. 

8.  Total  weight  of  water  fed  to  the  boiler Ibs. 

9.  Water  actually  evaporated,  corrected  for  moisture 

or  superheat  in  steam . . . .   " 

10.  Equivalent  water  evaporated  into  dry  steam  from 

and  at  212°  ,  .   " 

Hourly  Quantities. 

11.  Dry  coal  consumed  per  hour Ibs. 

12.  Dry  coal  per  sq.  ft  of  grate  surface  per  hour " 

13.  Water  evap.  per  hr.  corr.  for  quality  of  steam.  . .  .   " 

14.  Equivalent  evap.  per  hr.  from  and  at  212° " 

15.  Equivalent  evap.  per  hr.  from  and  at  212°  per  sq. 

ft.  of  water  heating  surface " 


174  THE  ELEMENTS  OF  STEAM  ENGINEERING 

Average  Pressures,  Temperatures,  etc. 

16.  Steam  pressure  per  gauge Ibs.  per  sq.  in. 

17.  Temperature  of  feed  water  entering  boiler degrees. 

18.  Temperature  of  escaping  gases  from  boiler "  • 

19.  Force  of  draft  between  damper  and  boiler ins.  of  water. 

20.  Percentage  of  moisture  in  steam  or  number  of  de- 

grees of  superheating per  ct.  or  deg. 

Horse-Power. 

21.  Horse-power  developed  (Item  14  -5-  34.5) .H.  P. 

22.  Builder's  rated  horse-power " 

23.  Percentage  of  builder's  rated  horse-power  devel- 

oped   per  cent. 

Economic  Results. 

24.  Water  apparently  evaporated  under  actual  condi- 

tions per  Ib.  of  coal  as  fired.     (Item  8  -f-  Item 
3) Ibs. 

25.  Equivalent  evaporation  from   and  at  212°   per 

pound  of  coal  as  fired.   (Item  9 -=- Item  3) .  . .  .   " 

26.  Equivalent  evaporation  from   and  at  212°   per 

of  dry  coal.    (Item  9  -f-  Item  5) " 

27.  Equivalent  evaporation  from   and  at   212°   per 

pound  of  combustible.      [Item  9  -f-  (Item  5 

-Item  6)] « 

(If  Items  25,  26,  and  27  are  not  corrected  for 
quality  of  steam,  the  fact  should  be  stated.) 

Efficiency. 

28.  Calorific  value  of  the  dry  coal  per  pound B.  T.  U. 

29.  Calorific  value  of  the  combustible  per  pound 

30.  Efficiency  of  boiler  (based  on  combustible) per  cent. 

31.  Efficiency  of  boiler,  including  grate  (based  on  dry 

coal)    

Cost  of  Evaporation. 

32.  Cost  of  coal  per  ton  of pounds  delivered  in 

boiler  room $ 

33.  Cost  of  coal  required  for  evaporating  1000  pounds 

of  water  from  and  at  212°  .  .  .  .  $ 


ECONOMY  OF  THE  STEAM  ENGINE  AND  BOILER          175 

155.  Weighing  the  Feed  Water. — The  apparently  simple  opera- 
tion of  weighing  the  feed  water  requires  the  utmost  care,  for  the 
omission  of  any  one  of  the  operations  attending  it  may  render  the 
results  of  the  test  valueless. 

The  arrangement  of  the  necessary  apparatus  for  a  boiler  test  is 
often  governed  by  local  conditions,  but  generally  there  are  two 
weighing  tanks  or  barrels  placed  on  scales,  the  water  supply  being 
connected  with  the  tanks.  The  capacity  of  these  tanks  may  vary 
±rom  5  to  15  cubic  feet — say  33  to  112  gallons,  or  312  to  937 
pounds.  These  tanks  should  be  labeled,  tank  No.  1  and  tank  No.  2, 
and  be  placed  above  the  level  of  a  third,  or  feeding,  tank  having  a 
capacity  somewhat  greater  than  either  of  the  two  weighing  tanks, 
so  that  the  contents  of  either  of  the  weighing  tanks  may  be  dis- 
charged into  it  before  it  is  entirely  emptied  without  danger  of 
overflowing. 

Preparatory  to  a  test  both  weighing  tanks  should  be  filled  from 
the  source  of  supply  and  their  gross  weights  recorded.  When  the 
level  of  the  water  in  the  feeding  tank  is  lowered  to  a  well-defined 
and  easily  observed  mark,  discharge  tank  Xo.  1  into  it,  and  note  the 
time  as  the  beginning  of  the  test.  The  weight  of  tank  No.  1  empty 
should  then  be  recorded  opposite  its  recorded  gross  weight.  As  the 
test  proceeds  tanks  Nos.  1  and  2  are  alternately  discharged  into  the 
feeding  tank  and  filled  from  the  source,  great  care  being  taken  that 
the  times  of  each  discharge  and  the  weights  of  the  tanks  full  and 
the  tanks  empty  be  recorded. 

To  end  the  test  the  level  of  the  water  in  the  boiler  is  brought  by 
the  feed  pump  or  injector  to  where  it  was  at  the  start,  the  discharge 
from  the  weighing  tank  being  so  regulated  that  the  level  of  the  water 
in  the  feeding  tank  shall  be  at  the  starting  mark  when  the  water  in 
the  boiler  is  at  its  required  level.  The  time  is  then  noted  as  the  end 
of  the  test.  It  is  obvious  from  this  that  all  the  water  used  during 
the  test  has  been  weighed  and  recorded. 

If  the  test  is  to  determine  the  steam  consumption  by  an  engine 
per  unit  of  power  the  weight  of  the  water  supplied  to  the  boiler 
furnishing  steam  exclusively  for  the  engine  is  measured  as  above 
described.  If  the  boiler  is  fed  by  a  pump  run  by  steam  from  the 
boiler  supplying  the  engine,  such  steam  must  be  accounted  for. 
This  can  readily  be  done  by  having  the  exhaust  from  the  pump  dis- 


CAl 


176 


THE  ELEMENTS  OF  STEAM  ENGINEERING 


charge  into  a  tank  resting  on  scales  and  containing  a  known  weight 
of  cold  water.  The  weight  of  the  tank  and  its  contents  can  then  be 
taken  at  intervals,  renewing  the  cold  water  when  the  temperature 
of  the  mixture  becomes  high  enough  to  vaporize.  If  the  boiler  is 
fed  by  an  injector  the  steam  used  for  operating  it  is  returned  to  the 
boiler  with  but  slight  loss. 

If  the  engine  be  of  the  surface  condensing  type  the  air  pump 
may  be  made  to  discharge  the  water  resulting  from  condensation 
into  a  filter,  from  which  it  may  be  drawn  into  weighing  tanks,  and 
then  fed  to  the  boiler  or  otherwise  disposed  of. 

156.  Record  of  a  Test. — The  following  form  of  keeping  a  record 
of  the  observations  made  during  a  test  is  the  one  used  by  the  stu- 
dents of  the  Baltimore  Polytechnic  Institute : 

MECHANICAL  LABORATORY, 
BALTIMORE  POLYTECHNIC  INSTITUTE, 

BALTIMORE,   MD. 
RECORD  OF  OBSERVATIONS  MADE  DURING  THE  TEST  OF 


Boiler, ,    19, 


Pressures. 

Temperatures. 

Fuel. 

Feed- 
water. 

Time. 

3 

, 

jj 

1 

6 

1 

I1 

Draugh 
gauge 

1 

M 

M 

Boiler-r 

i 

& 

i 

I 

J 

1 

5 

Pounds 

i 
e 

Pounds 
c.  ft. 





157.  Report  of  a  Test. — The  following  is  a  form  for  the  complete 
record  and  report  of  a  test : 


ECONOMY  OF  THE  STEAM  ENGINE  AND  BOILER 


177 


MECHANICAL  LABORATORY, 

BALTIMORE  POLYTECHNIC  INSTITUTE, 

BALTIMORE,  MD. 

Report  of  Boiler  Test  made  by 

Class ,  19 

Kind    of   Boiler , 

Manufactured  by 


Duration   of   trial      Hours 

^ 

Amount  used    Pounds 

5« 

Evaporated    dry   steam           Pounds 

Grate  surface,  length  ft., 
width  ft.,    Sq.  ft. 

p>  w 

Evap.  from  and  at  212°  Pounds 

Water-heating  surface       ....  SQ.  ft. 

PER  POUND  OF  FUEL 

K 

Superheating   surface                 Sq    ft 

*  Actual      .      Pounds 

r. 
% 
H 

Area    for    draught    (calorime- 
ter)                 Sq.  ft. 

Equiv.  from  and  at  212°.  .  .  .Pounds 
PER  POUND  OF  COMBUSTIBLE. 

3 

• 

Actual      Pounds 

Q 

Height    chimney      Ft. 

H 

Equiv.  from  and  at  212°.  .  .  .Pounds 

Ratio  heating  to  grate  surface. 
Ratio  air-space  to  grate  surface. 

i 

PER  SQ.  FT.  HEATING-SURFACE 
PER  HE. 

ri 

&> 

Actual      Pounds 

3 

D 

Steam-^auge                                Pounds 

H 

Equiv.  from  and  at  212°.  .  .  .Pounds 

to 

\      & 

Draught-gauge  Inches  water 

FROM  100°  F.  TO  70  POUND  BY  GAGE. 

Per  pound  of  fuel                     Pounds 

PH 

Absolute    steam-pressure.  .  .  .Pounds 

Per  pound  of  combustible       Pounds 

External   air                         Degrees  F 

Boiler-room     Degrees  F. 

PER  SQUARE  FOOT  OF  GRATE. 

Flue     Degrees  F. 

%  • 

Actual,  from  feed-water  tem- 

!      M 

Mg 

perature    Pounds 

Pn 

Feed-water                            Degrees  F 

$o 

Equiv.  from  and  at  212°  Pounds 

|l      -a 

Steam                                      Degrees  F 

o 

PER  SQ.  FT.  OF  WATER-HEATING 

!'  ^ 

5|H 

SURFACE 

Total   coal   consumed  Pounds 

>b 

Moisture    in    coal                .   Per  cent 

j 

Dry  coal  consumed                    Pounds 

Equiv.  from  and  at  212°.  .  .  .Pounds 

i,   p 

Total   refuse    dry                      Pounds 

On  basis  34y2  Ibs.  equiv.  evap. 

i'  &» 

. 

per  hour    HP 

£H 

Builder's  rating   H.P. 

w 

Dry  coal  per  hour  Pounds 

er's   rating    

« 
p 

0 

Combustible   per    hour  Pounds 
Combustible   per   square   foot 

Heat  generated  per  hour  B.T.U. 
Heat  absorbed  per  hour           B  T  U 

a 

of  grate    Pounds 

Efficiency   of   boiler                Per  cent 

1  S 

Dry   coal   per   square  foot  of 

Efficiency  of  furnace  Per  cent 

_! 
9 

Combustible   per   square   foot 
of  heating-surface              Per  cent 

fa 

Dry  coal  per   square  foot  of 
heating-surface    Per  cent 

Quality   of  steam  Per  cent 
Superheat                                   Degrees 

Total   water   used  Pounds 

J  «' 

1 

Total  water  used  (by  meter),  cu.  ft. 
Total  evap.,   dry  steam  Pounds 
Factor   of  evaporation    

Total  from  and  at  212°  Pounds 

*  Evaporation  from  temperature  of  feed-water  to  dry  steam  at  gauge-pressure ;  that  is,  it  is  the 
pparent  evaporation  corrected  from  calorimeter  test. 
12 


178  THE  ELEMENTS  OF  STEAM  ENGINEERING 

PKOBLEM. 

61.  The  pulley  of  a  dynamometer  is  26  inches  in  diameter.  The 
revolutions  of  the  engine  are  300  per  minute,  and  the  length  of  the 
beam  from  the  center  line  of  shaft  to  the  point  of  application  of 
the  pressure  on  the  scales  is  100  inches.  What  pressure  will  the 
scales  register  when  the  engine  is  developing  70  brake  horse- 
power? Ans.  147  pounds,  nearly. 


CHAPTEE  XIV. 
STEAM  BOILERS. 

158.  Classification  of  Boilers. — Steam  boilers  may  be  grouped 
into  three  classes,  viz. :   Stationary,  Locomotive,  and  Marine. 

The  stationary  class  includes  a  variety  of  forms,  the  most  im- 
portant being  the  horizontal  and  vertical  fire  tubular  and  the  sec- 
tional types. 

The  locomotive  boiler  in  universal  use  consists  of  a  rectangular 
tire  box  attached  to  a  cylindrical  shell  containing  a  number  of  tubes 
through  which  the  furnace  gases  pass  directly  to  the  smoke  pipe. 
This  boiler  possesses  great  steaming  capacity,  and  is  compact, 
durable,  and  of  moderate  cost.  It  is  not  infrequently  used  in  con- 
nection with  stationary  and  portable  engines. 

Marine  boilers  differ  in  form  and  setting  from  those  in  use  on 
land.  The  high  steam  pressures  demanded  by  modern  engineering 
conditions  have  completely  displaced  the  rectangular  type  of  boiler 
commonly  in  use  for  marine  purposes  as  late  as  the  year  1875.  The 
spherical  form  being  impossible,  the  cylinder  was  naturally  sug- 
gested as  the  most  practical  form  to  give  strength  to  boilers,  and 
the  type  known  as  the  cylindrical  return-fire-tubular  boiler  is  used 
largely  on  ships.  Special  types  of  the  Babcock  &  Wilcox  and  of  the 
Mclausse  sectional  boilers  have  lately  been  used  advantageously  for 
marine  purposes,  and  it  seems  likely  that  some  form  of  the  sectional 
water  tubular  boiler  will  eventually  displace  the  boiler  of  the  cylin- 
drical or  tank  form. 

159.  Sectional  Boilers. — This  type  of  water  tube  boiler  is  now 
very  largely  used  on  land,  and  is  passing  through  the  experimental 
stage  prior  to  its  eventual  adoption  for  marine  purposes.     A  sec- 
tional boiler  may  be  defined  as  one  in  which  the  contained  water  is 
divided  into  numerous  small  masses  connected  with  each  other  by 
passages  sufficiently  large  to  permit  free  circulation,  but  not  large 
enough  to  permit  so  sudden  a  release  of  pressure,  in  case  of  rupture 
in  one  of  the  sections,  as  to  cause  a  disastrous  explosion. 

There  are  many  varieties  of  sectional  boilers,  but  generally  speak- 


180  THE  ELEMENTS  OF  STEAM  ENGINEERING 

ing  they  may  be  said  to  consist  of  one  or  more  cylindrical  drums 
on  top  which  are  connected  by  a  series  of  tubes  with  a  drum  or 
drums  at  the  bottom.  The  heating  surface  is  made  up  almost 
entirely  of  the  tubes,  which  are  of  moderately  small  diameter  and 
therefore  very  strong,  though  made  of  thin  metal.  These  boilers 
may  be  made  light  and  of  great  power,  and,  by  reason  of  the  small 
volume  of  water  they  contain,  steam  may  be  raised  very  quickly, 
but  for  the  same  reason  variations  in  pressure  quickly  occur  in  cases 
of  unsteady  feed  or  irregular  firing. 

160.  Relative  Advantages  of  Different  Types  of  Boilers. — As 

already  stated,  the  Scotch  cylindrical  form  of  boiler  is  largely  in 
use  in  ocean  steamships  and  vessels  of  war,  and  it  is  regarded  as  a 
type  that  fills  the  requirements  of  such  service  in  a  fairly  satisfac- 
tory manner. 

In  small  craft,  such  as  torpedo  boats  and  yachts,  where  speed  is 
the  desideratum,  the  advantage  in  weight  possessed  by  the  water 
tube  boiler  has  rendered  the  use  of  some  one  of  its  forms  a  necessity. 

Professor  Durand,  in  comparing  the  water  tube  boiler  with  the 
boiler  of  the  Scotch  cylindrical  fire  tubular  type,  states  that  the 
weight  of  the  latter  without  water  is  usually  from  25  to  30  pounds 
per  square  foot  of  heating  surface,  while  that  for  the  lightest  type 
of  the  water  tube  boiler  ranges  from  12  to  20  pounds.  The  weight 
of  the  contained  water  per  square  foot  of  heating  surface  is  usually 
from  12  to  15  pounds  for  Scotch  boilers  and  from,  say,  1.5  to  3 
pounds  for  water  tube  boilers;  so  that  Scotch  boilers  with  water 
will  weigh  from  35  to  50  pounds  per  square  foot  of  heating  surface, 
while  the  weight  for  the  water  tube  boiler  will  be  from  13.5  to  23 
pounds.  It  seems,  however,  that  the  heating  surface  in  a  Scotch 
boiler  is  more  efficient,  square  foot  for  square  foot,  than  in  a  water 
tube  boiler,  so  that  it  is  customary  to  give  to  the  latter  type  10  to 
20  per  cent  more  heating  surface  for  equal  powers.  It  must  be 
remembered,  though,  that  the  water  tube  boiler  will  stand  forcing 
to  a  much  higher  degree  than  the  fire  tube  boiler.  With  the  latter 
supplying  steam  to  triple-expansion  engines  the  ratio  of  heating 
surface  to  I.  H.  P.  can  hardly  be  reduced  below  2,  while  with  the 
former  this  ratio  has  been  reduced  in  many  cases  to  1.5.  From  the 
nature  of  the  construction  of  the  water  tube  boiler  it  is  better 
adapted  to  stand  the  high  pressures  demanded  by  modern  practice, 
and  less  liable  to  disastrous  explosion.  From  one-quarter  to  one- 


STEAM  BOILERS  181 

half  hour  is  sufficient  time  in  which  to  raise  steam  in  a  water  tube 
boiler,  but  from  three  to  four  hours  should  be  taken  with  a  fire  tube 
boiler.  The  water  tube  boiler  has  the  additional  advantage  of  being 
portable  in  sections,  the  sections  being  readily  assembled  in  place 
upon  arrival  at  destination. 

It  may  be  said  to  the  disadvantage  of  the  water  tube  boiler  that 
it  requires  fresh  water  feed;  that  it  requires  a  uniform  feed,  due 
to  the  small  water  space  in  the  boiler  ;  that  it  cannot  readily  be  made 
in  units  larger  than  1000  horse-power,  while  double  that  is  not  un- 
common with  the  Scotch-  boiler  ;  and  that  a  rupture  of  a  tube  re- 
quires the  insertion  of  a  new  one,  necessitating  the  drawing  of  the 
fire  and  blowing  down  of  the  boiler. 

161.  To  show  the  necessity  for  a  steady  feed  for  the  water  tube 
boiler,  we  will  assume  a  boiler  having  50  square  feet  of  grate 
surface  and  burning  35  Ibs.  of  coal  per  hour  per  square  foot  of 
grate.  The  consumption  of  coal  would  then  be  35  X  50  =  1750 
Ibs.  per  hour.  With  a  consumption  of  1.7  pounds  of  coal  per 

I.  H.  P.  per  hour  would  mean  the  development  of  „  =  1029.5 
I.  H.  P.  Assuming  the  consumption  of  steam  per  I.  H.  P.  per 

-|  AOQ  5  y  1  4.  * 

hour  to  be  14  Ibs.,  an  evaporation  of  -        '?£       —  6.435  tons  of 


water  per  hour  would  be  necessary.     If,  when  the  water  is  at  the 
working  level,  the  weight  of  water  in  the  boiler  is  0.6435  ton,  the' 

boiler  would  have  to  be  filled     '         =  10  times  in  an  hour,  so  that 


a  cessation  of  the  feed  for  only  6  minutes  would  empty  the  boiler. 

162.  Boiler  Design.  —  The  problem  of  boiler  design  is  too  exten- 
sive for  present  consideration,  and  only  certain  of  its  features,  and 
their  inter-relation,  will  be  presented. 

There  is  no  similarity  in  the  process  of  designing  "  shell  "  and 
"  sectional  "  boilers.  The  sectional  boiler  seems  to  have  reached 
its  present  degree  of  perfection  by  a  pure  process  of  experiment,  for 
the  question  of  its  strength  involves  but  few  calculations.  All  its 
tubes  and  their  connections  have  much  greater  strength  than  that 
necessary  to  withstand  the  steam  pressure,  so  that  the  question  of 
design  involves  only  those  features  leading  to  the  production  of  a 
form  of  boiler  which  will  be  simple  in  management,  economical  in 
operation,  and  amply  protected  against  injury.  The  volume  of 


182 


THE  ELEMENTS  OF  STEAM  ENGINEERING 


steam  carried  by  a  sectional  boiler  is  so  small  compared  with  that 
carried  by  the  shell  boiler  that  the  question  of  circulation  is  of 
great  importance,  and  the  form  that  secures  the  freest  circulation, 
other  things  being  equal,  is  to  be  preferred. 

163.  The  design  of  the  cylindrical  boiler  involves  a  greater  num- 
ber of  calculations  than  can  be  here  attempted,  but  a  statement  of 
the  common  proportions  governing  the  problem  will  be  of  interest. 

Let  Fig.  52  represent  a  section  perpendicular  to  the  axis  of  a 
cylindrical  boiler  shell  of  internal  diameter  D  and  thickness  t,  both 
measured  in  inches.  If  p  denotes  the  internal  pressure  in  pounds 
per  square  inch,  then  the  radial  pressures,  each  equal  to  p,  above 
the  horizontal  diameter  are  balanced  by  those  below  it.  Consider 
a  very  small  arc  db,  of  length  x  and  one  inch  in  width,  as  acted  on 


by  the  radial  force  px.     The  vertical  component,  px  sin  a,  is  alone 
effective  in  producing  stress  in  the  sections  of  the  metal  at  the  ex- 


tremities   of   the   horizontal    diameter.     But   sin  a  =  —  ,    whence 

flS 

ac  =  eg  =  x  sin  a.  Hence  the  component  of  the  force  acting  on 
the  arc  db  which  produces  stress  in  the  metal  sections  at  the  ex- 
tremities of  the  horizontal  diameter  is  p  X  eg.  By  summing  up 
the  vertical  components  of  all  the  elemental  forces  acting  on  the 
upper  semicircle,  the  resultant  force  tending  to  produce  rupture  at 
the  metal  sections  is  pD. 

If  the  length  of  the  shell  be  L  inches,  this  resultant  force  be- 
comes pDL,  and  is  the  magnitude  of  the  force  tending  to  rupture 
the  shell  in  a  longitudinal  direction.  The  molecular  forces  which 


STEAM  BOILERS  183 

resist  rupture  are  measured  by  the  product  of  the  tensile  strength 
of  the  metal  and  the  areas  of  the  sections  made  by  the  longitudinal 
plane.  If  /  denotes  the  tensile  strength  of  the  metal  in  pounds 
per  square  inch,  then  the  resisting  force  is  2ftL.  At  the  point  of 
rupture  we  must  have  pDL  —  2ftL,  whence  pD  =  2ft. 

Considering  the  strength  of  the  transverse,  or  circular,  section  of 
the  shell,  the  stress  is  occasioned  by  the  pressure  on  the  two  ends. 
Eegardless  of  the  shape  of  the  ends,  the  pressure  tending  to  rupture 

the  shell  transversely  is   ^--  .     The  resisting  force  is  irDft,  and 

at   the   point   of   rupture   we   must   have  — r*-  =  irDft,    whence 

pD  =  4ft.  It  is  thus  seen  that  a  cylindrical  boiler  shell  is  twice 
as  strong  circumferentially  as  it  is  longitudinally.  The  tensile 
strength  of  boiler  steel  may  be  taken  as  65,000  Ibs.  per  sq.  inch, 
and  the  factor  of  safety  should  not  exceed  5. 

164.  The  first  quantity  usually  determined  in  designing  a  shell 
boiler  is  the  grate  surface,  and  the  recent  practice  shows  that  for 
each  square  foot  of  grate  surface  there  should  be  from  8  to  12 
I.  H.  P.  developed  by  the  engine,  depending  upon  the  nature  of 
the  draft  and  type  of  boiler,  the  water  tubular  type  requiring  the 
larger  grate  area. 

The  heating  surface  varies  from  25  to  35  square  feet  for  each 
square  foot  of  grate. 

The  coal  burned  per  square  foot  of  grate  surface  varies  from 
15  to  45  pounds  per  hour,  according  to  the  nature  of  the  draft, 
and  the  water  evaporated  per  pound  of  coal  is  reckoned  at  from 
6  to  10  pounds. 

The  sectional  area  of  the  tubes  is  taken  as  J  to  -f  the  area  of  the 
grate  surface,  and  the  sectional  area  over  the  bridge  wall,  and  of 
the  smoke  pipe,  from  ^  to  -J  of  the  grate  surface. 

The  volume  of  the  combustion  chamber  is  taken  as  from  3  to  4 
cubic  feet  per  square  foot  of  grate  surface,  and  the  volume  of  the 
steam  space  from  0.35  to  0.6  cubic  foot  per  I.  H.  P. 

165.  The  strength  of  a  furnace  tube  to  resist  external  pressure  is 
inversely  proportional  to  its  length  and  to  its  diameter,  and  before 
the  introduction  of  the  corrugated  furnace  it  was  the  practice  to 
make  the  furnace  in  sections,  flanged  at  the  ends  and  secured  together 


184  THE  ELEMENTS  OF  STEAM  ENGINEERING 

through  the  medium  of  the  Adamson  ring,  Fig.  53  (a)  or  of  t^ie 
Bowling  hoop,  Fig.  53  (6).  The  ring  or  hoop  gave  sufficient  stiff- 
ness to  the  furnace  and  was  universally  used  until  the  introduction 
of  the  corrugated  furnace. 


Figure  54  illustrates  the  Fox  corrugated  furnace,  which  is  very 
largely  used  in  boiler  construction.  It  is  an  extension  of  the 
principle  of  the  Bowling  hoop,  and  adds  immensely  to  the  strength 


of   the   furnace.      Its   strength   is    calculated   from    the    formula 

14,000£  .     , 

p  =   — h —  ,  in  which,  p  =  pressure  in  pounds  per  square  inch, 

t  =  thickness  in  inches,  and  d  =  the  least  outside  diameter  in 
inches.  Corrugated  furnace  tubes  have  no  longitudinal  joint,  being 
rolled  by  special  machinery  to  a  true  circular  form.  * 

166.  The  ends  of  the  Scotch  boiler  are  secured  by  means  of  stay 
rods  running  from  end  to  end,  screwing  into  the  heads,  and  fur- 
ther secured  by  nuts  inside  and  outside,  and  a  washer  outside  to 
increase  the  area  supported  by  the  brace.  The  back  end  of  the 
combustion  chamber  is  secured  to  the  back  end  of  the  boiler  by 
means  of  screw  stay  bolts. 

The  tubes  are  secured  into  the  tube  sheets  by  expanding  them 
and  beading  over  the  back  ends  as  a  protection  from  the  flame  in 
the  combustion  chamber.  The  Dudgeon  tube  expander  is  now 
universally  used.  If  no  stay  tubes  are  used,  both  the  front  and 
back  ends  of  the  ordinary  tubes  must  be  beaded  so  as  to  brace  the 
tube  sheets.  Stay  tubes  are  made  of  extra  heavy  material  and  are 
threaded  into  the  tube  sheets.  As  an  additional  protection  to  the 
stay  tubes  from  flame  they  are  usually  fitted  with  cast  iron  ferrules 
at  the  back  ends. 


STEAM  BOILERS 


185 


167.  The  flat  top  of  the  combustion  chamber  of  a  Scotch  boiler 
is  braced  with  girder  stays,  such  as  shown  in  Fig.  55.  It  is  a  case 
of  a  simple  beam  supported  at  the  ends,  with  concentrated  loads  at 
the  stay  bolts.  The  area  supported  by  each  girder  is  seen  from  the 
figure  to  be  28.5  X  7.5  =  213.75  square  inches,  and  as  the  pressure 
is  200  Ibs.  per  square  inch,  the  total  load  to  be  supported  is  42,750 
pounds.  Let  'P  denote  this  total  load;  -then  the  load  P'  at  each  of 

P  P 

the  three  stay  bolts  is  j.     The  reaction  at  each  support  is  -^  . 

The  maximum  bending  moment  will  be  at  the  middle  stay  bolt, 

,r        PL      PL      PL 

and  we  shall  have  Mmax  =  — -  —  -=-  =  — -  • 


T)  T  O  7" 

Hence  -~-  =  —  .     The  safe  working  stress  is  taken  as  8500  Ibs. 

D  C 

per  square  inch.     I  =  y~-  ,  c  =  ~  ,  and  L  =  28.5  inches. 


Then  42,750  x  28.5  =  8500  x  8* 


18xf 


X  28.5  =  85<M»«P. 


The  girders  are  arranged  in  pairs,  and  assuming  the  width  of  each 
as  J  inch,  we  shall  have  &  =  1.75  inches.     Then  for  the  depth  of 

742,750  X  28  5 

the   girder  we   shall   have,    d  =  J   Q,An  ^  -,  ~K-  =  9.05   inches, 

y     oouu  x  J-.  *  o 

which  agrees  with  the  depth  of  the  girder  as  shown. 

168.  The  shell  being  twice  as  strong  circumferentially  as  it  is 
longitudinally,  the  circumferential  joints  are  lapped  and  triple 
riveted,  while  the  longitudinal  joints  are  double-butt-strapped  and 
triple  riveted. 

When  flat  surfaces  are  to  be  stayed,  each  stay  must  stand  the 
pressure  on  the  area  it  supports,  so  the  sectional  area  of  the  stay 
must  be  such  that  its  tensile  strength  will  bear  the  strain  safely. 
If  a  denotes  the  pitch  of  the  stays,  d  the  diameter  of  the  stay,  /  the 


186  THE  ELEMENTS  or  STEAM  ENGINEERING 

safe  stress  of  the  metal,  and  p  the  pressure  per  square  inch  in 
pounds,  we  shall  have  pa2  =  -^~-  The  pitch  of  the  stays  of  flat 

surfaces  depends  largely  on  the  thickness  of  the  plate  to  be  stayed. 
If  the  pitch  be  too  great,  or  the  plate  too  thin,  there  is  danger  of 
the  plate  yielding  by  bulging.  Stay  rods  such  as  those  tying  the 
flat  ends  of  a  boiler  are  pitched  from  13  to  19  inches,  while  the 
screw  stay  bolts  tying  combustion  chambers  to  <*ach  other  or  to  the 
shell  of  the  boiler  are  pitched  from  7  to  8  inches. 

Sometimes  the  ends  of  cylindrical  boilers,  and  very  generally 
the  heads  of  steam  drums,  are  dished,  or  curved,  to  form  a  spherical 
surface,  in  order  that  they  may  retain  their  form  under  internal 
pressure ;  in  such  cases  no  bracing  is  required  for  the  heads. 

Sometimes  the  flat  ends  are  secured  to  the  cylindrical  shell  by 
gusset  stays,  which  consist  of  flat  plates  riveted  to  angle  irons 
secured  to  the  ends  and  shell.  Gusset  stays  are  also  used  at  times 
to  secure  the  combustion  chamber  to  the  back  end  of  the  boiler. 

169.  The  Eoberts  boiler  used  in  connection  with  the  mechanical 
laboratory  of  the  Baltimore  Polytechnic  Institute  is  a  good  speci- 
men of  the  water  tube  type  of  boiler. 

The  feed  water  from  the  pump  is  divided  at  the  boiler  into  two 
streams  by  a  T,  each  stream  entering  a  feed  coil  on  one  side  or  the 
other  of  the  drum  and  above  the  boiler  proper.  Passing  through 
the  horizontal  layers  of  these  coils,  it  is  then  delivered  into  the 
drum  above  the  water  line  in  order  that  any  steam  that  may  have 
formed  during  the  passage  through  the  coils  may  rise  to  the  top 
of  the  drum.  One  of  the  feed  coils  delivers  into  one  end  of  the 
drum  while  the  other  delivers  into  the  opposite  end. 

It  is  a  feature  of  this  boiler  that  the  course  of  the  water  through 
the  feed  coils  is  down,  thus  meeting  the  heated  gases  of  the  fire  as 
they  rise.  However  much  the  gases  may  be  cooled  by  contact  with 
the  lower  layers  of  the  coils,  they  meet  still  cooler  layers  as  they 
rise,  thus  establishing  the  most  favorable  conditions  for  the  trans- 
ference of  the  heat  of  the  gases  to  the  water. 

Under  the  action  of  gravity  the  water  now  flows  from  both  ends 
of  the  drum  through  T  connections  into  the  cross  pipes  and  thence 
through  the  large  downflow  pipes  into  the  side  pipes  below,  which 
are  on  a  level  with  the  fire  grate  (see  Fig.  56).  These  side  pipes 
are  tapped  for  the  upflow  coils,  through  which  the  water  now  rises 
again  to  the  drum.  It  will  be  seen  that  the  upflow  coils  rise  ver- 


STEAM  BOILERS 


187 


tically  from  the  side  pipes  and  then  cross  in  layers  under  the  drum 
and  directly  over  the  fire.  The  holes  in  the  side  pipes  for  the  up- 
flow  coils  are  "staggered"  so  that  the  spaces  between  the  cross 
layers  of  the  coils  directly  over  the  fire  are  only  one-half  as  wide 


Fig.  56. 

as  those  between  the  vertical  pipes  of  the  coils.  It  will  be  noted 
too  that  the  upflow  coils  leading  from  one  side  pipe  deliver  to  the 
drum  on  the  opposite  side.  The  feed,  now  arriving  at  the  drum 
for  the  second  time,  is  in  the  condition  of  damp  steam  and  bubbles 
through  the  water  in  the  drum  and  rises  to  enter  the  spray  pipe. 


188 


THE  ELEMENTS  OF  STEAM  ENGINEERING 


This  spray  pipe  runs  from  one  end  of  the  drum  to  the  other  and  is 
connected  to  the  heads  at  the  highest  points  possible.  It  has  a 
series  of  small  holes  drilled  in  its  center  line  along  the  top,  their 
purpose  being  to  prevent  water  entering  the  pipe  with  the  steam. 
The  steam  now  passes  from  both  ends  of  the  spray  pipe  into  the 
superheating  coils  which  lie,  one  on  each  side,  between  the  boiler 
and  its  inclosing  casing  and  above  the  fire  brick  of  the*  furnace. 


Fig.  57. 

The  steam  passes  to  the  bottoms  of  the  superheaters  and  thence  rises 
to  a  point  nearly  over  the  front  end  of  the  drum  where  a  T  con- 
nection enables  the  superheated  steam  to  be  led  to  the  safety  valve, 
engine,  etc. 

Figure  56  shows  the  drum,  the  large  cross  and  downflow  pipes  at 
front  and  rear,  the  side  pipes,  and  the  upflow  coils  leading  from 
them.  Near  the  top  of  the  front  head  of  the  drum  will  be  noticed 


STEAM  BOILERS  189 

the  L  for  connecting  the  spray  pipe  with  the  superheating  coil  on- 
the  left  side  of  the  boiler. 

A  perspective  view  of  the  complete  boiler  with  casing  removed 
is  shown  in  Fig.  57.  The  angular  pipe  on  the  right  side  leads 
from  the  rear  end  of  the  spray  pipe  to  the  superheating  coil  on  the 
right  side  of  the  boiler,  and  the  course  of  the  stream  through  the 
superheater  and  up  to  the  large  T  over  the  head  of  the  drum  can 
readily  be  traced. 

PROBLEMS. 

62.  The  steam  pressure  in  a  cylindrical  boiler  is  to  be  maintained 
at  210  pounds  per  square  inch.     Diameter  of  shell,  14  feet,  and 
the  efficiency  of  the  riveted  joints,  85  per  cent.     Using  a  factor  of 
safety  of  4,  it  is  required  to  find  the  necessary  thickness  of  shell. 

Ans.  1.25  inches. 

63.  The  Roberts  boiler  of  the  B.  P.  I.  has  21  sq.  ft.  of  grate  sur- 
race.     What  should  be  the  diameter  of  the  smoke  pipe? 

Ans.  22  inches. 

64.  A  Scotch  marine  boiler  is  to  supply  steam  for  an  engine  of 
820  I.  H.  P.,  the  boiler  to  use  forced  draft;   determine  the  follow- 
ing: (a)  Grate  surface.     (&)  Heating  surface,     (c)  Sectional  area 
of  tubes,     (d)  Sectional  area  over  bridge  wall,     (e)  Sectional  area 
of  smoke  pipe.     (/)  Volume  of  combustion  chamber,     (g)  Volume 
of  steam  space. 

Ans.  (a)  68.33  sq.  ft.  (b)  2050  sq.  ft.  (c)  '13.67  sq.  ft. 
(d)  11.39  sq.  ft.  (e)  11.39  sq,  ft.  (/)  273.32  cu.  ft. 
(g)  28.7  cu.  ft. 

65.  The  flat  top  of  the  combustion  chamber  of  a  Scotch  boiler 
is  3  feet  wide  and  2  feet  deep,  and  is  to  be  braced  with  girder  stays, 
the  boiler  pressure  being   180  Ibs.  per  sq.   inch.     Assuming  the 
width  of  the  stays  to  be  J-  inch  and  the  pitch  of  the  stay  bolts 
7  inches,  find  the  number  and  depth  of  the  girder  stays  and  the 
number  and  diameter  of  the  stay  bolts. 

Ans.  Five  pairs  of  girders,  6f  inches  deep ;  10  1^-inch  stay  bolts. 

66.  The  diameters  of  the  corrugations  of  a  furnace  flue,  from 
outside  to  outside,  are  39  inches  and  36  inches,  and  the  pressure 
to  be  carried  is  200  Ibs.  per  sq.  inch.     Find  the  thickness  of  the 
flue.  Ans.  T9g  inch. 


CHAPTER  XV. 
THE   STEAM   TURBINE. 

170.  The  most  important  modern  development  in  the  utilization 
of  heat  energy  by  mechanical  means  is  that  exemplified  in  the 
steam  turbine. 

In  the  reciprocating  steam  engine  the  motion  of  the  piston  is 
obtained  by  means  of  the  pressure  due  to  the  static  energy  of  the 
steam.,  whereas  in  the  steam  turbine  the  aim  is  to  convert  the  static 
energy  of  the  steam  into  kinetic  energy  and  then  to  apply  it  to  the 
rotating  parts  of  the  turbine. 

When  turbines  are  classified  according  to  the  direction  of  flow 
of  the  steam  there  are  two  classes : 

(a)  Radial-flow  turbines,  in  which  the  steam  flows  from  the 
center  to  the  circumference  or  from  the  circumference  to  the  center 
of  the  turbine  wheel. 

(.&)  Parallel-flow  turbines,  in  which  the  general  direction  of  the 
flow  of  the  steam  is  parallel  to  the  axis  of  the  wheel. 

171.  The  application  of  the  turbine  to  marine  propulsion  has 
attracted  the  attention  of  engineers,  and,  while  it  may  yet  be  said 
to  be  in  the  process  of  development,  it  cannot  be  regarded  as  ex- 
perimental from  the  fact  that  the  Allen  Line  has  contracted  to  con- 
struct a  turbine-driven  steamer  for  ocean  service  and  the  dinar d 
Company  is  to  add  to  its  fleet  two  turbine-driven  ships  of  60,000 
horse-power  each  and  to  attain  a  speed  of  25  knots  an  hour. 

The  field  of  usefulness  of  the  steam  turbine  in  other  directions 
is  extensive.  It  is  highly  successful  in  its  application  to  belt  trans- 
mission in  driving  machine  tools,  to  electric  generators,  to  blowers, 
and  to  the  operation  of  centrifugal  pumps.  It  is  particularly  suit- 
able to  the  last  named  application,  as  the  speeds  required  for  the 
best  efficiency  of  these  pumps  are  readily  obtained. 

The  small  turbines  of  one  power-shaft  of  the  De  Laval  type 
when  operating  centrifugal  pumps  are  quite  efficient  for  lifts  of 
from  15  to  150  feet,  the  larger  sizes,  with  two  power-shafts,  being 
used  for  lifts  of  from  40  to  300  feet.  For  greater  lifts  the  pump 
is  directly  connected  to  the  turbine  shaft,  the  pump  wheel  then 
revolving  with  a  velocity  of  from  10,000  to  30,000  revolutions  per 
minute.  The  pump  wheel  will  naturally  be  very  small  and  will 


THE  STEAM  TURBINE 


191 


not  produce  any  suction,  but  must  be  fed  by  another  pump  which 
is  connected  to  the  gear  shaft  and  running  at  a  considerably  re- 
duced velocity.  This  auxiliar}^  pump  sucks  the  water  and  presses 
it  into  the  high-speed  pump  wheel,  which  gives  the  high  pressure 
desired.  Pumps  of  this  type  have  been  made  for  lifts  up  to  a 
normal  head  of  850  feet  on  a  single  wheel  which,  at  a  decreased 
water  quantity,  can  go  up  to  1000  feet,  the  small  pump  giving  an 
efficiency  of  about  64  per  cent. 


Fig.  58. 

172.  Figure  58  is  a  representation  of  the  wheel  and  divergent  noz- 
zles of  the  De  Laval  turbine.  The  casing  through  which  the  nozzles 
enter,  and  which  encloses  the  wheel,  is  removed,  as  are  also  the 
outer  ends  of  some  of  the  vanes,  or  buckets,  in  order  that  the  general 
idea  of  the  course  and  action  of  the  steam  may  be  plainly  illustrated. 
The  nozzles  are  set  at  an  agle  of  20°  to  the  plane  of  the  wheel  j  and 
one  of.  them  is  shown  transparent  so  that  its  divergent  section  may 
be  shown.  The  complete  expansion  of  the  steam  takes  place  in  the 
nozzle.  It  enters  the  small  end  at  initial  pressure  and  is  discharged 
at  the  large  end  at  about  atmospheric  pressure  if  the  turbine  is  non- 
condensing,  and  at  a  pressure  corresponding  to  the  vacuum  if  con- 


192 


THE  ELEMENTS  OF  STEAM  ENGINEERING 


densing.  The  volume  of  the  nozzle  is,  therefore,  of  the  first  im- 
portance, as  the  entrance  orifice  must  be  of  sufficient  area  to  admit 
the  requisite  amount  of  steam,  and  at  the  discharge  end  the  area 
must  be  such  that  enables  the  complete  expansion  of  the  steam  to 
take  place  during  its  passage  through  the  nozzle.  The  proper  length 
of  the  nozzle  to  insure  a  regular  flow  of  steam  and  to  minimize 
friction  has  been  determined  experimentally. 


Pig.  59. 

During  the  expansion  of  the  steam  within  the  nozzle,  with  the 
consequent  fall  in  pressure,  its  heat  energy  is  converted  from  static 
to  kinetic,  so  that  the  issuing  jet  of  steam,  impinging  on  the 
crescent-shaped  vanes,  causes  the  turbine  wheel  to  revolve  with  a 
velocity  which,  in  practice,  reaches  1350  feet  per  second  for  the 
largest  powers  (300  H.  P.)  and  the  minimum  of  600  feet  per  sec- 
ond for  the  smallest  powers. 

173.  Figure  59  shows  the  working  parts  of  a  De  Laval  turbine. 
Mounted  on  the  turbine  shaft,  A,  is  the  turbine  wheel,  B,  fitted 


THE  STEAM  TURBINE  193 

with  a  series  of  vanes  around  its  circumference.  These  vanes  are 
lettered  F  in  Fig.  60.  The  pinion,  C,  is  made  solid  with  the  shaft 
and,  when  in  position,  engages  the  gear  wheel,  H,  mounted  on  the 
power  shaft,  7.  The  pinion  bearings,  E  and  D,  and  the  power-shaft 
bearings,  J  and  L,  are  in  two  parts,  as  shown.  The  ball  bearing, 
F  (marked  8  in  Fig.  61)  bears  most  of  the  weight  of  the  wheel  and 
is  self-adjusting,  being  held  in  its  seat  by  a  cap  and  spring,  as  shown. 
The  flexible  bearing,  G  (marked  T  in  Fig.  61)  is  free  to  move  in 
response  to  any  oscillations  of  the  turbine  shaft  and  is  kept  seated 
by  a  cap  and  spring;  its  object  is  to  prevent  the  escape  of  steam 
when  running  non-condensing,  or  the  admission  of  air  when  run- 
ning in  connection  with  a  condenser. 


Fig.  60. 

174.  Figure  60  is  a  section  showing  the  divergent  nozzle.     The 
initial  steam  from  the  chamber,  D,  enters  the  nozzle  through  the 
valve,  H,  and  after  completing  its  expansion  within  the  nozzle,  im- 
pinges with  great  velocity  on  the  vanes,  F.     The  number  of  these 
nozzles  varies  from  1  to  15,  according  to  the  power  of  the  machine, 
and  any  of  them  may  be  put  out  of  action  by  closing  their  inlet 
valves,  thus  regulating  the  power  developed. 

175.  Figure  61  is  a  plan  section,  and  Fig.  62  an  elevation  section 
of  a  30  H.  P.  De  Laval  turbine  dynamo.     The  steam  entering 
through  the  governor  valve,  (7,  Fig.  62,  passes  into  the  chamber,  D, 
Fig.  61,  and  thence  to  the  nozzles.     After  the  transfer  of  its  kinetic 
energy  to  the  vanes  the  steam  passes  into  the  chamber,  6r,  and  is 
finally  exhausted  through  H.     The  governor,  0,  Fig.  62,  is  held  in 
the  end  of  the  power  shaft,  L,  by  means  of  a  tapered  shank,  and 
operates  by  means  of  the  bell-crank,  'P,  and  the  governor  valve,  C. 
The  power  shaft  is  coupled  to  the  dynamo  at  M. 

13 


194  THE  ELEMENTS  OF  STEAM  ENGINEERING 

176.  In  the  construction  of  the  wheel  of  the  De  Laval  turbine  the 
problem  of  successfully  resisting  the  stresses  produced  by  the  cen- 
trifugal force  arising  from  the  high  velocity  presented  itself.  As 


Fig.  61. 


THE  STEAM  TURBINE 


195 


196 


THE  ELEMENTS  OF  STEAM  ENGINEERING 


these  stresses  vary  with  the  square  of  the  velocity  it  was  evident 
that  a  wheel  in  the  form  of  a  flat  disc  would  burst  at  the  normal 
speed  required  for  the  operation  of  the  turbine,  so  the  form  of  the 
wheels  shown  in  Fig.  63  was  devised  to  meet  the  requirements.  As 


Fig.  63. 

shown  by  the  sections,  the  wheel  is  heavy  at  the  boss  and  tapers  in 
section  toward  the  peripheral  ring  to  which  the  vanes  are  attached. 
The  centrifugal  force  at  the  boss  is  relatively  small  and  is  a 
maximum  at  the  periphery.  The  gradual  increase  in  section  area 
from  the  boss  to  the  periphery  provides  an  excess  of  material  that 
becomes  available  for  the  support  of  the  highly  stressed  section  at 


THE  STEAM  TURBINE 


197 


the  periphery.  The  vanes  are  inserted  into  milled  slots  in  the 
peripheral  ring,  and  the  centrifugal  force  due  to  their  velocity  loads 
the  ring  to  an  amount  equal  to  the  stresses  produced  in  the  body  of 
the  wheel.  The  section  of  the  wheel  immediately  under  the  ring 
is  purposely  made  small,  and  so  designed  that  fracture  will  take 
place  at  that  section  should  the  speed  approximate  double  that  of 
the  normal.  A  fracture  of  that  description  would  break  the  ring 
into  such  small  pieces  that  no  damage  to  the  wheel  case  would  result, 
and,  the  vanes  then  gone,  the  action  of  the  steam  on  the  wheel 
would  immediately  cease. 


The  wheels  of  the  larger  sizes  are  made  solid  in  body  in  order 
to  give  the  maximum  of  supporting  material  at  the  boss,  and,  in 
such  cases,  the  shaft  is  made  in  two  parts  and  secured  to  the  wheel 
as  shown  in  Fig.  63. 

177.  Figure  64  is  an  approximate  three-quarter  size  representa- 
tion of  two  of  the  200  vanes  of  a  300  H.  P.  De  Laval  wheel,  one  of 
them  shown  in  mid-section  by  a  plane  perpendicular  to  the  axis  of-' 
rotation.  The  vanes  are  drop-forged  from  moulded  steel  and  are 
fitted  into  milled  slots  in  the  rim  of  the  wheel.  Fig.  65  is  a  front 
view  of  a  vane,  and  Fig.  66  a  section  on  AB  of  Fig.  64. 


198  THE  ELEMENTS  OF  STEAM  ENGINEERING 

The  diameter  of  a  300  H.  P.  De  Laval  wheel  is  about  30  inches, 
and  the  weight  of  one  vane  about  0.7  ounce. 

178.  Owing  to  the  lack  of  uniform  density  of  metals  it  is  impos- 
sible to  produce  a  wheel  which  will  balance  about  its  axis  of  forma- 
tion, and  in  consequence  very  serious  vibrations  would  result  at  the 
high  speeds  attained  by  the  De  Laval  wheel  should  no  provision  be 
made  for  their  prevention. 

The  length  and  lightness  of  the  shafts  on  which  these  wheels  are 
mounted  admit  of  such  flexibility  that  when  the  so-called  critical 
speed  is  reached  the  wheel  is  permitted  to  choose  an  axis  of  rotation 
through  its  center  of  gravity,  with  the  consequent  absence  of  vibra- 
tions and  perfect  smoothness  of  running.  The  high  rotary  speeds 
of  these  shafts  admit  of  slender  dimensions,  those  of  the  smaller 
powers  being  but  f  inch  in  diameter,  while  that  of  the  maximum 
of  300  H.  P.  is  but  If  inches.  There  is  always  sufficient  clearance 
between  the  wheel  and  its  casing  to  allow  for  the  small  displace- 
ment of  the  wheel  incident  to  its  choosing  its  axis  of  rotation. 

179.  In  the  De  Laval  turbine  the  steam  acts  on  a  single  set  of 
vanes,  and  since  it  can  be  shown  that  for  maximum  efficiency  the 
vane  speed  should  be  about  half  the  velocity  of  the  steam  jet,  it 
follows  that  the  use  of  even  moderately  high  steam  pressures  pro- 
duces such  excessive  velocities  for  the  vanes,  with  the  consequent 
stresses  due  to  centrifugal  action,  that  a  limit  is  put  on  the  efficiency 
by  the  strength  of  the  material  available  for  construction.     It  is 
also  a  condition  for  efficiency  with  the  De  Laval  wheel  that  the  jet 
shall  leave  the  vanes  at  a  minimum  velocity,  thus  securing  the 
maximum  transference  of  energy. 

This  latter  condition  may  be  dispensed  with  in  a  steam  turbine 
if  it  be  so  arranged  that  the  steam  acts  on  a  series  of  sets  of  vanes, 
their  angles  and  velocities  being  such  that  a  portion  of  the  energy 
of  the  steam  is  given  up  to  each  set.  This  is  the  basic  principle  of 
the  Westinghouse-Parsons,  and  of  some  other  turbines,  by  means  of 
which  high  rotational  speeds  are  avoided  and  the  ability  to  develop 
high  powers  secured. 

180.  Figure  67  illustrates  the  action  of  the  steam  on  the  vanes  of 
a  Westinghouse-Parsons  turbine,  the  dotted  lines  showing  the  course 
of  the  steam  and  the  horizontal  arrows  the  direction  of  rotation  of 
the  moving  vanes. 


THE  STEAM  TURBINE  199 

The  fixed  vanes  project  radially  inward  from  the  enveloping 
casing.  Mounted  on  the  shaft,  which  is  co-axial  with  the  casing, 
is  a  hollow  cylindrical  drum,  from  the  outer  surface  of  which  the 
moving  vanes  project  radially  outward.  The  entering  steam  first 
encounters  a  ring  of  fixed  vanes,  from  which  it  is  deflected  so  as 
to  impinge  in  such  a  direction  on  the  adjoining  ring  of  moving 
vanes  as  to  give  to  them  a  rotary  impulse.  Leaving  this  ring  of 
moving  vanes  the  steam  next  encounters  another  ring  of  fixed  blades 
and  is  by  them  deflected  in  the  proper  direction  for  impact  on  the 
next  ring  of  moving  blades.  The  force  rotating  the  moving  vanes 
is  the  double  one  of  the  impact  of  the  entering  steam  and  the 
reaction  of  the  effluent  steam.  The  passages  through  the  vanes  are 
of  constantly  increasing  volume,  corresponding  to  the  increased 
volume  of  the  steam.  This  increase  in  volume  of  the  passages  is 


^  I    I 


F/XE0 
MOy/HG 

Ft 4.  67. 

obtained  by  increasing  the  heights  of  the  vanes  until  the  mechanical 
limit  of  such  increase  is  reached,  when  the  diameter  of  the  turbine 
is  abruptly  increased,  thus  allowing  for  another  progression  in  the 
heights  of  the  vanes. 

It  is  thus  seen  that  the  passage  of  the  steam  through  the  many 
passages  of  gradually  increasing  volume  permits  a  gradual  expan- 
sion of  the  steam  and  the  conversion  of  its  energy  into  velocity, 
until,  finally,  the  desired  pressure  and  temperature  is  reached  for 
exhaustion  into  the  atmosphere  or  into  the  condenser. 

181.  Figure  68  is  a  longitudinal  section  through  a  Westinghouse- 
Parsons  turbine,  and  is  described  in  a  paper  by  Mr.  Francis  Hodg- 
kinson  as  follows : 

"  The  steam  enters  at  the  governor  valve  and  arrives  at  the  cham- 
ber, A,  and  passes  out  to  the  right  through  the  turbine  blades,  even- 
tually arriving  at  the  exhaust  chamber,  B.  On  the  left  of  the 
steam  inlet  are  shown  revolving  balanced  pistons,  C,  C^,  (72,  one 


200 


THE  ELEMENTS  OF  STEAM  ENGINEERING 


THE  STEAM  TURBINE  201 

corresponding  to  each  of  the  cylinders  in  the  turbine  which,  accord- 
ing to  size,  may  be  1,  2,  3,  or  4  in  number.  The  steam  at  A  presses 
against  the  turbine  and  goes  through  doing  work.  It  also  presses 
in  the  reverse  direction,  but  cannot  pass  the  piston,  (7,  but  at  the 
same  time  the  pressure,  so  far  as  the  steam  at  A  is  concerned,  is 
equal  and  opposite,  so  that  the  shaft  is  not  subjected  to  any  end 
thrust. 

"  The  pressure  at  D  is  equal  to  that  at  E  by  reason  of  the  balance 
port,  F;  similarly,  so  far  as  the  steam  pressure  at  E  is  concerned, 
there  is  no  end  thrust.  This  same  fact  also  applies  to  G. 

"  The  areas  of  the  balance  pistons  are  so  arranged  that  no  matter 
what  the  load  may  be,  or  what  the  steam  pressure  or  exhaust  pres- 
sure may  be,  the  correct  balance  is  preserved  and  the  shaft  has  no 
end  thrust  whatever. 

"At  H  is  shown  a  thrust  bearing  which,  however,  has  no  thrust 
to  take  care  of,  but  serves  to  maintain  the  correct  adjustment  of  the 
balance  pistons. 

"At  K  is  a  pipe  connecting  the  back  of  the  balance  pistons  at  L 
with  the  exhaust  chamber,  to  insure  the  pressure  at  this  point  being 
exactly  the  same  as  that  of  the  exhaust. 

"At  J  are  shown  the  bearings.  They  are  unique  in  construction. 
The  bearing  proper  is  a  gun  metal  sleeve,  which  is  prevented  from 
turning  by  a  loose  fitting  dowel.  Outside  of  this  are  three  concen- 
tric tubes  having  a  small  clearance  between  them.  This  clearance 
fills  up  with  oil  and  permits  a  vibration  of  the  inner  shell,  at  the 
same  time  restraining  same.  The  shaft  therefore  revolves  about  its 
axis  of  gravity,  instead  of  the  geometric  axis,  as  would  be  the  case 
were  the  bearings  of  every-day  construction.  The  journal  is  thus 
permitted  to  run  slightly  eccentric,  according  as  the  shaft  may  be 
out  of  balance.  This  form  of  bearing  in  a  very  remarkable  man- 
ner performs  the  functions  of  De  Laval's  slender  flexible  shaft. 
But  in  this  case  the  shaft  is  built  as  rigidly  as  possible,  so  is  not 
liable  to  crystallization,  which  would  result  in  eventual  rupture. 

"  The  bearings  have  ample  surface,  are  continuously  lubricated 
under  pressure,  and  it  has  been  found  in  practice  that  they  do  not 
wear. 

"At  R  is  shown  a  flexible  coupling  by,  means  of  which  the  power 
of  the  turbine  is  transmitted.  In  small  sizes  the  two  shafts  have 
a  square  cut  on  the  ends,  the  coupling  itself  somewhat  loosely  fitting 


202  THE  ELEMENTS  OF  STEAM  ENGINEERING 

over  these.  In  larger  sizes  it  is  generally  a  modification  of  this 
arrangement. 

"  The  governor  gear  and  oil  pumps  generally  receive  their  motion 
by  means  of  a  worm  wheel  gearing  into  a  worm  cut  on  the  outside 
of  the  coupling. 

"At  N  is  an  oil  reservoir  into  which  drains  all  the  oil  from  the 
bearings.  From  there  it  runs  into  the  pump,  M,  to  be  pumped 
up  to  the  chamber,  0,  where  it  forms  a  static  head  which  gives  a 
continuous  pressure  of  oil  to  the  bearings.  The  pump  is  single- 
acting  of  the  simplest  possible  construction,  that  will  not  become 
deranged.  The  oil  runs  in  by  gravity,  so  that  it  is  unlikely  to  fail 
to  continue  pumping. 

"A  by-pass  valve  is  provided,  shown  at  iP,  which  admits  high- 
pressure  steam  by  means  of  port  Q  to  the  steam  space  E.  By  open- 
ing this  valve  as  much  as  60  per  cent  overload  may  be  obtained, 
and  in  the  case  of  turbines  operating  condensing,  full  load  may  be 
obtained  should  the  condenser  be  at  any  time  inoperative,  due  to 
any  cause,  and  the  turbine  allowed  to  exhaust  into  the  atmosphere." 

182.  In  comparison  with  the  reciprocating  engine  it  may  be  said 
of  the  steam  turbine : 

(1)  It  is  more  simple  in  construction. 

(2)  It  is  more  permanent,  since  there  is  the  complete  absence  of 
reciprocating  parts,  and  the  bearings  so  completely  adjusted  and 
lubricated  as  to  almost  preclude  the  possibility  of  wear. 

(3)  The  size  and  weight  per  horse-power  is  less  than  that  for  the 
steam  engine. 

(4)  The  consumption  of  steam  per  horse-power  is  less  than  for 
the  simple  reciprocating  engine  and  about  the  same  as  that  for  the 
triple-expansion  engine. 

(5)  It  permits  the  use  of  steam  superheated  to  the  highest  de- 
gree, since  there  is  an  absence  of  the  necessity  for  internal  lubri- 
cation. 


PART  II 

TRANSMISSION  OF  POWER  BY 
BELTS  AND  GEAR-WHEELS 


CHAPTEE  I. 
BELTING. 

1.  The  transmission  of  power  and  of  motion  by  means  of  belts 
running  over  pulleys  is  an  important  and  familiar  mechanical  con- 
trivance. It  is  practically  noiseless,  and  may  be  used  for  trans- 
mitting power  through  a  distance  as  great  as  30  feet  without 
intervening  support.  The  principal  disadvantage  is  that  due  to  the 
slip  occasioned  by  the  freedom  of  the  belt  to  slip  on  the  pulley, 
rendering  the  transmission  not  so  positive  as  that  through  the 
medium  of  gear  wheels;  but  this  disadvantage,  in  cases  where 
mechanisms  at  rest  are  suddenly  thrown  into  gear,  becomes  an 
advantage  in  preventing  shocks.  Motion  may  be  transmitted  by 
belting  between  shafts  making  various  angles  with  each  other,  the 


necessary  provision  being  that  the  advancing  side  of  the  belt 
be  at  right  angles  to  the  shaft  it  approaches,  while  the  receding  side 
may  deviate  in  any  direction.  . 

2.  Figure  1  is  a  side  elevation  and  end  view  of  a  "  quarter  turn  " 
in  power  transmission  by  means  of  a  belt  and  pulleys,  the  shafts 
being  at  right  angles.  It  will  be  noted  that  the  belt  in  passing 
from  pulley  A  approaches  pulley  B  in  a  direction  at  right  angles  to 
the  axis  of  5,  and  in  passing  from  B  it  approaches  A  in  a  direction 
at  right  angles  to  the  axis  of  A,  as  will  be  seen  from  the  end  view. 
It  will  be  noted,  too,  that  a  plane  that  is  tangent  to  the  face  of  one 
pulley  and  perpendicular  to  the  axis  of  the  other  cuts  the  other 
pulley  in  the  middle  of  its  face. 


206     TRANSMISSION  OF  POWER  BY  BELTS  AND  GEAR-WHEELS 

3.  No  definite  rule  can  be  given  for  the  distance  between  shafts 
that  are  to  be  connected  with  each  other  by  belts.     The  rule  that 
ten  times  the  diameter  of  the  smaller  pulley  is  an  advantageous 
distance  is  not  of  general  application,  for  much  depends  on  the 
judgment  of  the  engineer  in  applying  general  principles  to  meet  the 
conditions  that  confront  him.    For  narrow  belts  running  over  small 
pulleys,  fifteen  feet  is  a  good  average  distance,  and  for  larger  belts 
and  larger  pulleys,  the  maximum  distance  is  about  thirty  feet.    In 
any  event  the  distance  should  be  such  as  to  permit  a  gentle  sag  to 
the  belt  when  it  is  in  motion. 

4.  Experiment  has  shown  that  belts  tend  to  run  on  the  highest 
part  of  pulleys,  and  in  order  to  keep  them  central  on  the  rims  a 
convexity  amounting  to  about  one-twenty-eighth  of  the  breadth  is 
given  to  the  surface  of  the  rims. 


5.  The  belt  connecting  two  pulleys  may  be  open  or  crossed,  the 
resulting  rotation  of  the  pulleys  being  in  the  same  or  in  opposite 
directions  respectively. 

6.  It  is  clear  that,  in  the  transmission  of  motion  from  one  pulley 
to  another,  the  outer  surface  of  the  rim  of  each  pulley  will  have  the 
same  velocity  as  the  belt;  so  if  we  denote  by  Dt  and  Z>2  the  diame- 
ters of  the  driver  and  follower  respectively,  and  by  N^  and  iV2  their 
revolutions  per  minute,  we  shall  have : 

Speed  of  outer  surface  of  driver  rim  =  7rP1AT1 ,  and 
Speed  of  outer  surface  of  follower  rim  =  irD2N2 , 
and  since  each  of  these  is  equal  to.  the  speed  of  the  band,  we  have 

vD^  —  irD2N2 ,  whence  :&  ss  & .     That  is,  the  velocities  of  a 

-Aj  i/2 

pair  of  pulleys  are  inversely  as  their  diameters. 


BELTING  207 

7.  When  motion  is  transmitted  from  one  shaft  to  another  by 
belting,  one  or  more  shafts  intervening,  as  in  Fig.  2,  we  shall  have 

&a*-§*9&±*2*9  and  ~6  =  5s.  Taking  the  product  of  these 
JVi  D2  Jvs  D4  -A5  D6 

equations,  member  by  member,  and  remembering  that  N2  =  N3 

and  'N±  —  N5)we  have  ^?  =  ^1 X  -^  X  ^5 .     That  is,  the  ratio  of 
iVj       £)2       .L>4      .L/6 

the  speed  of  the  last  pulley  to  the  speed  of  the  first  pulley  is  equal 
to  the  continued  product  of  the  ratios  of  the  diameters  of  each  pair 
of  pulleys  taken  in  order. 

EXAMPLE  I. — The  pulley  on  the  shaft  of  a  steam  engine  is  30 
inches  in  diameter,  from  which  a  belt  passes  to  a  pulley  20  inches  in 
diameter  on  a  shaft  in  a  room  above;  a  belt  passes  from  another 
20-inch  pulley  on  this  shaft  to  one  of  8  inches  in  diameter  on  a 
third  shaft;  an  18-inch  pulley  on  the  third  shaft  is  belted  to  a 
6-inch  pulley  on  the  spindle  of  a  dynamo:  find  the  speed  of  the 
dynamo  when  the  engine  is  making  100  revolutions  per  minute. 

-a-  Speed  of  Dynamo      30^20^.18    ,,       , 

Here,  -f — a — yw — -. —  =  ^  X  -^  X  -*-  >  therefore. 

Speed  of  Engine       20      86 

30     20     1 8 
Speed  of  Dynamo  =  —  X  —  x  -^X 100  =  11 25  revolutions  per  minute. 

8.  The  thickness  of  a  belt  may  have  a  very  appreciable  effect  on 
the  velocity  ratio  of  two  pulleys.     While  the  belt  is  in  contact  with 
the  pulley  its  inner  surface  is  in  compression  and  its  outer  surface 
in  tension,  but  the  neutral  surface  midway  between  is  of  constant 
length.     It  follows,  then,  that  the  velocity  of  the  surface  of  the  belt 
in  contact  with  the  pulley  is  less  than  the  velocity  of  the  neutral 
surface,  and  that  the  true  velocity  ratio  of  two  pulleys  connected 
by  a  belt  is  obtained  by  increasing  the  diameters  of  the  pulleys  by 
an  amount  equal  to  the  thickness  of  the  belt. 

If  the  thickness  of  the  belts  in  the  above  example  were  -fa  inch, 
and  the  thickness  be  taken  into  consideration,  we  shall  have : 

30  3     20  ^    18i 
Speed  of  dynamo  =  — ^x—^X— i^X  100 

*4  4  4 

=  1083.76 


m          QQ 
OU 

revolutions  per  minute.     The  effect  has  been  to  decrease  the  calcu- 
lated speed  of  the  dynamo  by  1125  ^^83'76  =  0.0367  or  by  3.67  f. 


208     TRANSMISSION  OF  POWER  BY  BELTS  AND  GEAR-WHEELS 

9.  Frictional  Resistance  between  a  Belt  and  a  Pulley. — Figure  3 
represents  two  pulleys  connected  by  a  belt.  With  the  pulleys  sta- 
tionary the  belt  is  strained  over  them  so  that  the  initial  tension 
will  insure  against  slipping.  If,  now,  a  force  tends  to  turn  the 
driving  pulley,  D,  in  the  direction  of  the  arrow,  the  lower  part  of 
the  belt  will  be  stretched  so  that  its  tension  will  be  increased.  The 


tension  in  the  upper  part  will  be  decreased  an  equal  amount.  Each 
element  of  the  belt  in  contact  with  the  pulley  assists  the  action  of 
the  tension  in  the  slack  side  of  the  belt  in  resisting  the  tension  in 
the  tight  side,  and  therefore  the  tension  in  that  part  of  the  belt  in 
contact  with  the  pulley  varies  at  every  point.  When  the  difference, 
TI  —  T2,  becomes  sufficient  to  overcome  the  resistance  to  motion 


in  the  driven  pulley,  F,  its  rotation  begins.  The  difference  in  ten- 
sion in  the  two  sides  of  the  belt  is  the  amount  of  friction  between 
the  belt  and  the  pulley,  and  is  the  measure  of  the  driving  force.  If 
the  tensions  were  equal  their  moments  about  the  center  of  the  driven 
pulley  would  be  equal  and  there  would  be  no  rotation,  since  there 
would  be  equal  turning  efforts  in  opposite  directions. 


BELTING  209 

Let  TI  and  T2,  Fig.  4,  denote  the  tensions  in  the  tight  and  slack 
sides  of  the  belt,  respectively.  Suppose  that  da  be  a  very  small 
part  of  the  central  angle  a  subtended  by  the  arc  of  contact  of  the 
belt. 

Considering  the  equilibrium  of  the  very  small  arc  subtending  da, 
we  will  assume  the  tensions  at  its  extremities  to  be  T  and  T  +  dT. 
If  the  resulting  reaction  between  the  belt  and  pulley  rim  due  to 

these  tensions  be  denoted  by  R,  we  shall  have  R  =  2T  sin  ~  =  Tda, 

A 

since  sin  —  and  —  —  are  very  approximately  the  same. 

<>  a 

Since  slipping  is  about  to  take  place,  dT  is  the  measure  of  the 
friction  over  the  small  arc  considered  ;  whence  dT  —  p,R  =.  fiTda  , 
in  which  p  is  the  coefficient  of  friction. 

j  rp 

We  have  then,  for  the  small  arc  considered,  -=-  =  pda,  and 

/»TI  ^T7          /»a 

for   the   whole   arc    of    contact  I       -,w-  =     I    u.da,    from    which 

JTZ      *         t/o 

rp  rp 

log  Tl  —  log   T2  =  /xa,   or   log  -=i  —  //.a;  whence  -^  =  e*a,  where 

*J  *I 

^  —  2.718,  the  base  of  the  Naperian  system  of  logarithms. 


We  shall  then  have,  Iog10      ±    —  pa  Iog10  2.718  =  0.4343/>ta. 

v  •*•-/ 
This  equation  shows  that  the  ratio  of  the  tensions  depends  only 

upon  the  coefficient  of  friction  and  the  angle  at  the  center  subtended 
by  the  arc  of  contact  of  the  belt,  and  is  independent  of  the  diameter 
of  the  pulley.  In  the  above  equations  a  is  expressed  in  circular 
measure. 

10.  The  results  just  obtained  hold  for  a  rope  making  several 
turns  about  a  post.  Thus,  in  the  equation  log  (  ~A  —  0.4343/m  ,  if 
n  denotes  the  number  of  turns  of  the  rope,  then  a  =  2-rrn,  and  we 
shall  have,  log(-^\—  0.4343  X  2^  ;  whence,  log  TI  —  log  T2 
4-2.729n/x. 

EXAMPLE  II.  —  A  rope  makes  three  turns  about  a  post,  one  end 
being  pulled  with  a  force  of  25  Ibs.  The  coefficient  of  friction 
being  0.4,  what  is  the  greatest  force  that  can  be  resisted  at  the 
other  end  ? 

Here,  T2  =  25  ,  n  =  3  ,  and  p  —  0.4. 

Then,  log  T^  =  log  25  +  2.729  X  1.2  ,    whence  2\  =  47,000  Ibs. 
14 


210    TRANSMISSION  OF  POWER  BY  BELTS  AND  GEAR-WHEELS 

11.  Transmission  of  Power  by  Belts.  —  The  friction  between  the 
belt  and  pulley  limits  the  amount  of  power  that  can  be  transmitted. 
When  overloaded,  a  belt  will  slip  rather  than  break,  therefore,  at 
the  point  of  slipping,  we  shall  have: 


Horse-power  transmitted  =  ^-^^  ,  in  which  F  =  T±  —  T2  in 

00,000 

pounds,  and  V  the  velocity  of  the  belt  in  feet  per  minute  ;  therefore, 

(T-T)V      7\(l-p-U       r,(l-->.-)r 
Horse-power  —  kz*  _  *J     —       \          2i  /      _  e    ' 

33,000  33,000  33,000 

Substituting  the  value  of  e,  and  letting  /*  =  0.35  and  a  =  180°  =  TT 
in  circular  measure,  we  shall  have  : 


Horse-power  -  ~          -    - 

33,000          '    33,000  * 

TI  may  be  taken  as  80  Ibs.  per  inch  width  of  single-ply  belt,  which 
makes  allowance  for  laced  joints.  If  w  =  width  of  belt  in  inches, 
then  TI  =  Sow  ,  and  the  equation  for  the  horse-power  becomes  — 


Horse-power  —     '  —  -         for  single-ply  belting,  and  for 


double-ply  we  shall  have  :  Horse-power  —  ™      . 

600 

These  two  approximate  equations  are  easily  remembered,  and  are 
useful  for  quickly  estimating  power. 

The  linear  velocity  of  belts  ranges  from  3000  to  6000  feet  per 
minute. 

12.  Creeping  of  Belts.  —  The  driving  side  of  a  belt  is  necessarily 
stretched  more  than  the  slack  side,  and  in  consequence  the  driving 
pulley  receives  a  greater  length  of  belt  than  it  gives  off  to  the  fol- 
lowing pulley;   therefore,  the  speed  of  the  driving  side  is  a  trifle 
greater  than  that  of  the  slack  side.     But  the  speed  of  the  rim  of  a 
pulley  is  the  same  as  that  of  the  belt  it  receives;    therefore,  the 
speed  of  the  rim  of  the  driver  will  be  somewhat  greater  than  that  of 
the  rim  of  the  follower,  and,  as  a  consequence,  the  speed  of  the 

follower  will  be  less  than  that  given  by  the  formula  —  ^2-  =  -^-. 

j\l        DI 

This  difference  between  the  speeds  of  the  rims  of  the  driver  and 
follower  is  known  as  creep,  or  slip,  and  amounts  to  about  2  per  cent. 

13.  Strength  of  Belting.  —  The  ultimate  strength  of  leather  used 
for  belting  varies  from  3000  Ibs.  to  5000  Ibs.  per  square  inch  of 


BELTING 


section,  and  the  strength  of  the  laced  joint  may  be  taken  as  one- 
third  that  of  the  solid  leather.  Taking  a  factor  of  safety  of  5,  the 
safe  working  tension  with  a  laced  joint  may  be  taken  from  200  to 
330  pounds  per  square  inch  of  section.  Single-ply  belts  range  in 
thickness  from  f$  inch  to  T5-g-  inch,  and  the  width  must  be  made 
sufficient  to  withstand  the  tension.  The  safe  tension  in  pounds 
per  inch  of  width  ranges  from  50  to  100,  according  to  the  thickness 
and  the  safe  stress  of  section  of  the  material. 

14.  Centrifugal  Action  in  Belts. — The  stress  in  the  rim  of  a  fly- 
wheel or  of  a  pulley,  due  to  centrifugal  action,  may  be  treated  in 
a  manner  similar  to  the  stresses  in  a  cylindrical  boiler  shell,  if  the 
effects  of  the  arms  be  neglected. 

In  this  instance  the  radial  force,  instead  of  being  the  pressure  of 

the  steam  per  square  inch,  is         —(see  p.  385),  in  which  r  is  the 

mean  radius  of  the  rim  in  feet;  and,  as  in  the  case  of  the  boiler 
shell,  the  resultant  centrifugal  force  producing  stresses  in  the 
sections  of  the  rim  at  the  extremities  of  a  diametral  plane  is 

W*)P  9  Wit2 

^LL.  X2r=  -:L£Z_  (see  p.  182).  This  force  is  balanced  by  the 
tensions  in  the  rim  at  the  two  sections,  so  that  the  whole  tension  in 


the  rim  is  —     -  in  pounds  per  square  inch. 

In  precisely  the  same  manner,  that  part  of  a  belt  embracing  a 
pulley  is  subjected  to  a  stress  due  to  centrifugal  action,  and  at  high 
speeds  part  of  the  tension  in  the  belt  is  expended  in  a  tendency  to 
lift  the  belt  as  it  passes  over  the  surface  of  the  pulley,  thereby 
decreasing  the  normal  pressure  on  the  pulley.  In  belt  speeds  in 
excess  of  50  feet  per  second  the  effect  of  this  centrifugal  action 
should  be  determined,  and  the  width  of  the  belt  increased  accord- 
ingly. 

One  linear  foot  of  leather  one  square  inch  in  section  weighs  0.43 
lb.,  so  that  one  foot  of  belting  a  square  inches  in  sectional  area  will 
weigh  0.43a  Ibs. 


The  effect  of  centrifugal  action  on  the  belt  is  then     -  ,  in 

which  v  is  expressed  in  feet  per  second.  The  length  of  a  foot  of 
belting  is  taken  because  the  velocity  is  expressed  in  feet  per  second, 
and  the  sectional  area  of  the  belt  is  taken  in  square  inches  because 
the  stress  is  expressed  in  pounds  per  square  inch. 


212     TRANSMISSION  OF  POWER  BY  BELTS  AND  GEAR-WHEELS 

In  all  cases  where  the  effect  of  centrifugal  action  in  belting  is  to 
be  considered,  the  real  tensions  are  to  be  taken  as  Tl  -j — : 

t/ 

andT2  +     QA3av*  .     It   will   be   noted   that   the    driving    force, 

}j 

T±  —  T2,  is  not  affected  by  the  centrifugal  action. 

EXAMPLE  III. — Find  the  width  of  belt  necessary  to  transmit 
10  horse-power  to  a  12-inch  pulley  so  that  the  greatest  tension  may 
not  exceed  40  pounds  per  inch  of  width  when  the  pulley  makes  1500 
revolutions  per  minute,  the  weight  of  the  belt  per  square  foot  being 
1.5  Ibs.,  the  coefficient  of  friction  0.25,  and  the  arc  of  contact  of 
the  belt  180°. 

0.43 


Here,  if  t  denotes  the  thickness  of  the  belt,  we  have :  144£  X 

=  1.5  ,  whence  t  =  0.29  inch. 

Then  v  =  *  X  12.29  X  1500  _  g(U3  ^  ^  gec(m(L 

-L<y  X  OU 

(ffi  rp\    y 

We  have,  Horse-power  =  ^    l         2; — ,  whence 

00,000 


12 


We  have,  S=:  e**,  whence  T^  =  Tze^=  T2  X  2.718°-257r=2.19T2 . 

Then,  2.19T2  =  68.37  +  T2 ,  whence  T2  =.  57.46,  and 

TI  —  68.37  +  57.46  =  125.83  Ibs. 
The  effect  of  centrifugal  action  increases  the  tension   by  the 

amount  °-43gt;' .     If  w  denotes  the  width  of  the  belt  in  inches,  then 

i 
a  =  tw  =  0.29w  ,  and  we  shall  have : 

0.43  X  0.29w  X  80.43  X  80.43 
—  =  ±#y.oo  -t-  -  —     n — 

g  32.2 

=  125.83  +  25.05w. 

But  from  the  conditions  of  the  problem  the  tension  must  not  exceed 
40w  ;   hence  40z/;  =  125.83  -(-  25.05w  ,  whence  w  =  8.4  inches. 

PEOBLEMS. 

1.  Two  pulleys,  15  inches  and  37.5  inches  in  diameter  respec- 
tively, are  connected  by  a  belt.  If  the  former  makes  500  revolu- 
tions in  10  minutes,  how  many  turns  will  be  made  by  the  other  in 
25  minutes?  Ans.  500. 


BELTING  213 

2.  Sketch  an  arrangement  of  four  pulleys  with  belts  for  driving 
a  fan  at  1500  revolutions  per  minute  from  a  shaft  making  200  revo- 
lutions per  minute,  giving  the  diameters  of  the  pulleys  to  be  used. 

3.  An  engine  shaft,  making  n  revolutions  per  minute,  carries  a 
56-inch  pulley  which  drives,  by  means  of  a  belt,  a  36-inch  pulley  on 
the  line-shaft.     The  line-shaft  carries  another  pulley,  42  inches  in 
diameter,  which  is  belted  to  a  24-inch  pulley  on  a  counter-shaft. 
Another  pulley  on  the  counter-shaft  is  48  inches  in  diameter  and 
is  belted  to  a  14-inch  pulley  on  the  spindle  of  a  dynamo.     Find  the 
number  of  revolutions  made  in  a  minute  by  the  dynamo  spindle. 

Ans.  9.33n. 

4.  The  flywheel  of  an  engine  is  28  inches  in  diameter  and  is 
belted  to   a  pulley  20   inches  in  diameter  on   another  shaft.     A 
20-inch  pulley  on  this  second  shaft  is  belted  to  a  10-inch  pulley  on 
a  third  shaft,  which  carries  an  18-inch  pulley,  which,  in  turn,  is 
belted  to  a  6-inch  pulley  on  the  spindle  of  a  dynamo.     Find  the 
speed  of  the  dynamo  when  the  engine  is  making  90  revolutions  per 
minute.     If  the  belt  thickness  of  three-sixteenths  of  an  inch  be 
considered,  find  the  per  cent  of  loss  in  speed  of  the  dynamo. 

Ans.  756  revolutions ;  3.18  per  cent. 

5.  The  shaft  of  a  high-speed  engine,  which  is  making  300  revo- 
lutions per  minute,  carries  a  14-inch  pulley,  over  which  passes  a  belt 
to  a  20-inch  pulley  on  another  shaft.     A  10-inch  pulley  on  this 
shaft  drives  a  20-inch  pulley  on  a  third  shaft  carrying  a  6-inch 
pulley  which  is  to  be  belted  to  the  spindle  of  a  machine  so  that  the 
revolutions  of  the  spindle  may  be  45  per  minute.    Find  the  diameter 
of  the  pulley  on  the  spindle.  Ans.  14  inches. 

6.  The  pulley  on  an  engine  shaft  is  5  feet  in  diameter  and  makes 
100  revolutions  per  minute.     The  motion  is  transmitted  from  this 
pulley  to  the  main  shaft  by  a  belt  running  on  a  pulley,  the  difference 
in  tensions  between  the  tight  and  slack  sides  of  the  belt  being  115 
pounds.     What  is  the  work  done  per  minute  in  overcoming  the  re- 
sistance to  motion  of  the  main  shaft?  Ans.  180,714  ft.  Ibs. 

7.  A  belt  having  a  linear  velocity  of  350  feet  per  minute  trans- 
mits 5  H.  P.  to  a  pulley.     Find  the  tension  on  the  driving  side,  sup- 
posing it  to  be  double  that  on  the  slack  side.  Ans.  943  Ibs. 


214     TRANSMISSION  OF  POWER  BY  BELTS  AND  GEAR-WHEELS 

8.  A  pulley  3  feet  6  inches  in  diameter,  and  making  150  revolu- 
tions per  minute,  drives  by  means  of  a  belt  a  machine  which  absorbs 
7  H.  P.     What  must  be  the  width  of  the  belt  so  that  its  greatest 
tension  shall  be  70  pounds  per  inch  of  width,  it  being  assumed  that 
the  tension  in  the  driving  side  is  twice  that  in  the  slack  side  ? 

Ans.  4  inches. 

9.  Find  the  H.  P.  that  may  be  transmitted  by  a  belt  8  inches  wide 
and  passing  over  a  20-inch  pulley  on  the  shaft  of  an  engine  which 
makes  350  revolutions  per  minute.     The  angle  at  the  center  sub- 
tended by  the  arc  of  contact  of  the  belt  is  160°,  and  the  coefficient 
of  friction  is  0.4.     The  tension  on  the  driving  side  is  not  to  exceed 
80  Ibs.  per  inch  of  width.  Ans.  23.94. 

10.  A  belt  laps  150°  round  a  pulley  of  3  feet  diameter,  making 
130  revolutions  per  minute;    the  coefficient  of  friction  is  0.35. 
What  is  the  maximum  pull  on  the  belt  when  20  H.  P.  is  being  trans- 
mitted and  the  belt  is  just  on  the  point  of  slipping? 

Ans.  900  Ibs. 

11.  A  belt  is  to  transmit  2  H.  P.  from  a  pulley  12  inches  in  diam- 
eter on  a  shaft  making  160  revolutions  per  minute.     Find:   (1) 
The  tensions  on  the  driving  and  on  the  slack  sides  of  the  belt  when 
the  belt  embraces  half  the  pulley  rim  and  the  coefficient  of  friction 
is  0.3.     (2)  The  width  of  the  belt  when  the  thickness  is  0.25  inch, 
the  safe  working  tension  being  320  Ibs.  per  square  inch  of  section. 

Ans.  215.15  Ibs.;   83.85  Ibs.;   2.69  inches. 

12.  What  H.  P.  will  a  belt  8  inches  wide  transmit  over  an  18-inch 
pulley  making  300  revolutions  per  minute,  the  weight  of  1  sq.  ft. 
of  the  belt  being  1.29  Ibs.,  the  coefficient  of  friction  0.3,  the  central 
angle  embraced  by  the  belt  160°,  and  the  tension  per  inch  of  width  of 
belt  not  to  exceed  80  Ibs?     Take  the  weight  of  a  foot  of  belting 
1  sq.  inch  in  section  as  0.43  Ib.  Ans.  15.8. 

13.  A  leather  belt  is  required  to  transmit  2  H.  P.  from  a  shaft 
running,  at  80  revolutions  per  minute  to  a  shaft  running  at  160. 
Find  the  stresses  in  the  belt,  assuming  that  the  smaller  pulley  is 
12  inches  in  diameter,  and  that  the  ratio  of  the  tensions  in  the 
tight  and  slack  sides  of  the  belt  is  2.25  :  1.     Then  find  the  width 
of  belt,  taking  the  working  stress  at  100  Ibs.  per  inch  of  width. 

Ans.  236.3  Ibs.;   105  Ibs.;  2.36  inches. 


BELTING  215 

14.  Find  the  width  of  belt  necessary  to  transmit  10  H.  P.  to  a 
pulley  12  inches  in  diameter  so  that  the  greatest  tension  may  not 
exceed  40  Ibs.  per  inch  of  width  when  the  pulley  makes  1500  revo- 
lutions per  minute,  the  weight  of  the  belt  per  sq.  ft.  being  1.5  Ibs., 
and  the  coefficient  of  friction  0.25.     The  weight  of  a  cubic  inch  of 
belting  may  be  taken  as  0.0358  lb.,  and  the  arc  of  contact  of  the  belt 
as  180°.     (The  effect  of  the  thickness  of  the  belt  and  of  the  cen- 
trifugal action  of  the  belt  must  be  taken  into  consideration.) 

Ans.  8.4  inches. 

15.  An  8-inch  belt,  traveling  over  a  30-inch  pulley  making  174 
revolutions  per  minute,  transmits  18  H.  P.     The  ratio  of  the  ten- 
sions in  the  tight  and  slack  sides  of  the  belt  is  3.06  :  1,  the  arc  of 
contact  160°,  and  the  maximum  tension  allowed  per  inch  width  of 
belt  80  Ibs.     Taking  the  velocity  of  the  neutral  surface  of  the  belt 
as  the  true  velocity,  it  is  required  to  find:  (1)  The  thickness  of  the 
belt.     (2)  The  coefficient  of  friction.  Ans.  0.25  inch;  0.4. 

16.  A  driving  shaft,  making  100  revolutions  per  minute,  carries 
a  pulley  22  inches  in  diameter,  from  which  a"  belt  communicates 
motion  to  a  12-inch  pulley  on  a  counter-shaft.   On  the  counter-shaft 
is  also  a  cone  pulley  having  steps  of  8,  6,  and  4  inches  in  diameter, 
which  gives  motion  to  another  cone  pulley,  of  equal  steps,  on  a 
lathe  spindle.     Sketch  the  arrangement  in  side  and  end  elevations, 
and  find  the  greatest  and  least  speeds  at  which  the  lathe  spindle  can 
revolve.  Ans.  366.66;  91.66. 

17.  Determine  the  H.  P.  that  may  be  transmitted  by  a  belt,  6 
inches  wide  and  0.25  inch  thick,  running  at  a  speed  of  60  feet  per 
second,  the  tension  in  the  slack  side  being  0.45  of  that  in  the  tight 
side  of  the  belt,  the  stress  allowed  being  280  pounds  per  sq.  inch 
of  section.     To  what  extent  does  the  effect  of  centrifugal  action 
reduce  the  power  transmitted?  Ans.  20.87;  17.2  per  cent. 

18.  A  weight  of  8000  pounds  is  suspended  from  one  end  of  a 
rope.     How  many  turns  of  the  rope  must  be  taken  around  a  circular 
beam  fixed  horizontally,  in  order  that  a  man,  who  can  pull  with  a 
force  of  250  pounds,  may  keep  the  rope  from  slipping,  supposing 
the  coefficient  of  friction  to  be  0.2?  Ans.  2.75. 

19.  By  taking  3  turns  of  a  rope  about  a  post,  and  holding  back 
with  a  force  of  180  pounds,  a  man  just  keeps  the  rope  from  slipping. 


216    TRANSMISSION  OF  POWER  BY  BELTS  AND  GEAR-WHEELS 

The  coefficient  of  friction  being  0.2,  find  the  weight  supported  at 
the  other  end  of  the  rope.  Ans.  7810  Ibs. 

20.  It  is  required  to  transmit  16  H.  P.  from  a  pulley  20  inches 
in  diameter  by  means  of  a  belt  which  embraces  only  two-ninths  of 
the  circumference  of  the  pulley.  Find  the  tensions  of  the  two 
parts  of  the  belt  when  slipping  is  just  prevented,  and  the  width  of 
belt  required,  the  thickness  of  the  belt  being  three-eighths  of  an 
Jnch,  the  safe  working  stress  300  Ibs.  per  sq.  inch,  the  speed  of  the 
pulley  120  revolutions  per  minute,  and  the  coefficient  of  friction 
0.35.  Ans.  2174.2  Ibs.;  1333.8  Ibs.;  19.32  inches. 


CHAPTER  II. 
WHEELS  IN  TRAIN. 

15.  The  circles  in  Fig.  5  represent  the  pitch  circles  of  two  toothed, 
or  spur,  wheels  in  gear.  The  pitch  circles  of  two  wheels  in  gear 
are  circles  which  appear  to  roll  upon  each  other  and  which  pass, 
approximately,  through  the  middle  of  the  elevation  of  the  teeth. 
They  may  be  regarded  as  the  outlines  of  two  discs  which  roll  to- 
gether by  the  friction  at  their  circumferences. 

There  can  be  no  slipping  of  one  pitch  circle  over  the  other,  owing 
to  the  teeth;  therefore,  the  same  length  of  circumference  of  each 
must  pass  over  the  point  of  contact  in  a  given  time.  If  D  and  d 
represent  the  diameters  of  the  large  and  small  wheels  respectively, 
and  N  and  n  the  number  of  their  revolutions  per  unit  of  time,  then 
-n-DN  =  irdn  ,  or  DN  =  dn  . 


The  teeth  on  both  wheels  are  of  the  same  size,  and  their  number 
will  be  proportional  to  the  diameters  of  the  wheels,  and  we  may 
therefore  write  TN  —  tn,  in  which  T  and  t  are  the  numbers  of 
teeth  on  the  large  and  small  wheels  respectively. 

16.  The  pitch  of  the  teeth  is  the  distance  measured  on  the  cir- 
cumference of  the  pitch  circle  between  the  centers  of  two  consecu- 
tive teeth.     The  diameter  of  the  pitch  circle  =  2—y  in  which  p  is 
the  pitch,  and  t  the  number  of  teeth. 

17.  The  formulas  for  belt  pulleys  hold  for  toothed  wheels,  for  the 
belt  performs  the  same  office  as  the  teeth — it  causes  the  circum- 
ference of  both  pulleys  to  move  over  the  same  distance  in  the  same 
time. 


218     TRANSMISSION  OF  POWER  BY  BELTS  AND  GEAR-WHEELS 

18.  The  usual  arrangement  of  a  train  of  toothed  wheels  in  a 
mechanism  is  shown  in  Fig.  6,  where  two  wheels  unequal  in  size  are 
placed  on  each  axis,  except  the  first  and  last,  and  where  the  larger 
wheel  of  any  pair  gears  with  the  next  smaller  wheel  in  the  train. 
On  the  first  and  last  axes  there  is  but  one  wheel. 

19.  Let  Tlt  T2,  T3,  and  Nlf  N2,  Ns  denote  the  number  of  teeth 
in  the  driving  wheels  of  each  pair  in  gear  and  their  number  of 
revolutions  in  a  period  of  time  respectively;   let  tlf  t2,  ts,  and  nt, 
n2,  nB  be  like  representations  for  the  follower  wheels  of  each  pair. 

From  what  has  been  shown,  we  shall  have : 

T^  =  t^ ,  T2N2  =  t2n2 ,  and  TSNZ  =  *8n, . 
Multiplying  these  equations,  member  by  member,  we  have : 
2\  X  T2  X  T3  X  # ±  X  N2  X  N*  =  *!  X  t,  X  *3  X  n,  X  n2  X  fia . 


Since  the  first  follower  and  the  second  driver  are  fixed  on  the 
same  spindle,  we  have,  ni  =  N2,  and  likewise,  n2  =  JV3 . 
Therefore,  N±  X  Z\  X  T2  X  Ts  =  ns  X  ^  X  t«  X  *3  - 

That  is : 

The  number  of  revolutions  of  the  first  driver,  multiplied  by  the 
continued  product  of  the  numbers  of  teeth  in  all  the  drivers,  is 
equal  to  the  number  of  revolutions  of  the  last  follower,  multiplied 
by  the  continued  product  of  the  numbers  of  teeth  in  all  the 
followers. 

If,  in  Fig.  6,  N±  =  16  per  minute,  TI  =  72  ,  T2  =  40  ,  Tz  =  36  , 
£±  —  24 ,  tz  =  20 ,  ts  =  18  ,  we  shall  have  by  substituting  in  the 
general  equation, 

16  X  72  X  40  X  36  =  ^3  X  24  X  20  X  18 , 

whence,  nz  —  192,  or  the  last  follower  will  make  192  revolutions 
while  the  first  driver  is  making  16. 

20.  The  ratio  of  the  number  of  revolutions  of  the  last  wheel  of  a 
train  in  a  given  time  to  the  number  of  revolutions  of  the  first  wheel 


WHEELS  IN  TRAIN  219 

in  the  same  time  expresses  the  Velocity  Katio,  or  the  Value,  of  the 
train,  and  will  be  denoted  by  e. 

To  find  e  in  the  above  example,  we  have : 

=  **.  ==  -192  =-  12  =     72X40X36 
A!        16    "      1"  ""  24X20X18 

Product  of  the  number  of  teeth  in  all"  the  drivers. 
~  Product  of  the  number  of  teeth  in  all  the  followers. 

21.  When  any  number  of  wheels  are  in  gear,  no  two  of  them 
being  on  the  same  axis,  as  in  Fig.  7,  the  combination  is  the  equiva- 
lent only  of  a  single  pair  of  wheels,  viz.,  the  first  wheel  and  the 
last  wheel;  the  intervening  wheels  simply  transfer  the  motion  and 
determine  the  direction  of  rotation  of  the  last  wheel.  If  the  num- 
ber of  idler  wheels  intervening  between  the  first  and  last  wheels  be 


odd,  the  direction  of  rotation  of  the  first  and  last  wheels  will  be  the 
same;  if  even,  the  rotation  will  be  in  opposite  directions. 

22.  Generally  speaking,  the  part  of  a  machine  to  which  the  motive 
force  is  applied  is  called  the  Driving  End,  and  the  part  at  which 
the  resistance  is  overcome,  or  at  which  the  useful  work  is  done,  is 
called  the  Follower  End,  or  the  Load  End. 

Movement  of  Driving  End  .       ,,   ,   , 

The  ratio,  _ ,  _  .„ -^ — ,    -,    ,    is  called  the 

Movement  of  Follower,  or  Load,  End 

Velocity  Ratio  of  the  machine. 

Load 


We  have  seen  that  Mechanical  Advantage  =  .  . 

Driving  Force 

hence, 

Mechanical  Advantage  _          Load  X  Movement  of  Load  End 

Velocity  Ratio         ~~  Driving  Force  X  Movement  of  Driving  End 

Useful  Work  of  Machine  __  •MeciL(in:cai 

== — ?j — -= =-; — y^ — =r= — j—. aunfitwHVAM 

Work  Supplied  to  Machine 


220      TRANSMISSION  OF  POWER  BY  BELTS  AND  GEAR-WHEELS 


23.  Transmission  of  Power  by  Gear  Wheels.  —  In  the  different 
mechanisms  employing  a  train  of  toothed  wheels  for  the  transmis- 
sion of  power,  either  the  movement  of  a  small  power,  P,  through  a 
comparatively  great  distance,  is  utilized  in  overcoming  a  much 
greater  resistance,  W,  through  a  much  smaller  distance;  or,  con- 
versely, the  movement  of  a  large  power,  P,  through  a  small  dis- 
tance is  utilized  in  making  a  smaller  resistance,  W,  move  through 
a  much  greater  distance.  The  desiderata  in  the  two  cases  are 
power  and  velocity  respectively,  the  underlying  mechanical  prin- 
ciple being  that,  what  is  gained  in  power  is  lost  in  speed,  and 
conversely. 

The  power,  P.,  may  be  applied  by  hand  to  the  end  of  a  lever  or 
to  the  crank  pin  of  an  engine,  the  lever  or  crank  to  be  rigidly  con- 
nected to  the  first  axis  of  the  train. 


24.  Two  toothed  wheels  of  unequal  size,  fixed  on  the  same  axis, 
is  the  mechanical  equivalent  of  a  lever  with  unequal  arms,  and 
therefore  modifies  the  power  that  may  be  transmitted. 

The  toothed  wheels  B  and  (7,  Fig.  8,  are  fixed  on  the  same  axis,  c. 
The  driving  power,  P,  is  applied  through  wheel  A  tangentially  to 
the  pitch  surface  of  the  teeth  in  contact  at  5,  the  tendency  being 
to  turn  wheels  B  and  C  in  a  contra-clockwise  direction  about  axis  c. 
The  action  of  P  is  resisted  by  the  reaction  P^  of  the  load,  applied 
tangentially  to  the  pitch  surface  of  the  teeth  of  the  wheels  C  and  D 
in  contact  at  d,  with  a  tendency  to  turn  the  wheels  B  and  C  in  a 
clockwise  direction  about  axis  c.  With  these  opposite  turning  ten- 
dencies, we  shall  have,  by  the  principle  of  the  lever  : 


,  whence,  -£  =  4  = 
'   P         cb 


Radius  of  wheel  B 
Number  of  teeth  in  wheel  C 
Number  of  teeth  in  wheel  B  " 


WHEELS  IN  TRAIN 


221 


25.  The  double  purchase  wheelwork  of  Fig.  9  represents  that 
commonly  applied  to  hoisting  machinery,  such  as  cranes.  The 
numbers  attached  to  the  wheels  and  pinions  indicate  the  number  of 
teeth  they  contain.  The  length  of  the  lever  handles  being  18  inches, 
and  the  diameter  of  the  drum  20  inches,  the  weight  that  can  be 
raised  by  the  application  of  40  pounds  at  each  of  the  handles  may 
be  calculated  by  the  principle  of  the  lever,  as  follows : 

Since  the  radii  of  wheels  are  proportional  to  the  number  of  teeth 


in  the  wheels,  we  may  denote  by  ISx  and  120z  the  radii  of  the  first 

pinion  and  the  last  wheel  of  the  train  respectively. 

Let  R  =  Length  of  lever  handles, 
"     r  =  Eadius  of  drum, 
"    P!,  P2,  P3  =  Tangential  pressures  at  the  points  of  contact  of 

teeth  in  gear. 

Then,  by  the  principle  of  the  lever,  we  have : 

P  x  R  =  P±  X  18ar  (1) 

P±  X  80  =  P2  X  40  (2) 

P2  X  120  =  P3  X  20  (3) 

P3  X  120z  =  W  X  r  (4) 


222    TRANSMISSION  OF  POWER  BY  BELTS  AND  GEAR-WHEELS 

Multiplying  these  equations,  member  by  member,  we  have  : 
P  X  R  X  80  X  120  X  120z  =  W  X  r  X  18a?  X  40  X  20. 
w  P  _  r      1S#  X  40  X  20         r  _  r  .      1 

C6'  ~W  •~5'80X120Xl20i  ~"%*       -~EXW 


w= 


80(40  +  40)18 


r  10 

In  the  above  calculations  the  effect  of  friction  and  the  diameter 
of  the  rope  have  been  neglected.  Friction  would  be  likely  to  re- 
duce the  result  as  much  as  30  per  cent,  and  the  effective  drum 
radius  would  be  the  radius  of  the  drum  plus  the  radius  of  the  rope. 

26.  The  tangential  pressures,  P19  P2,  P3  may  be  found  by  means 
of  equations  (1),  (2),  (3),  and  (4),  if  the  pitch  of  the  teeth  be 
known. 

Thus,  suppose  the  pitch  of  the  teeth  to  be  1.25  inches.     Then, 

Radius  of  first  pinion  =  A^X_1?        45 


4rr 

Radius  of  last  wheel  =  !•»*  X  120  =  WO  § 

2;:  2rr 

From  equation  (1)  we  have: 

80  X  18  =  P!  X  |jp  whence,  P±  =  402.3  Ibs. 

From  equation  (2)  we  have: 

402.3  X  80  =  40  X  P2 ,  whence,  P2  =  804.6  Ibs. 
From  equation  (3)  we  have: 

804.6  X  120  =  20  X  P3 ,  whence  P3   =  4827.6  Ibs. 
These  results  would  indicate  that  the  teeth  of  the  wheels  should 
be  made  stronger  as  the  drum  shaft  is  approached.     This  could  be 
done  by  using  a  coarser  pitch  for  the  last  pair  of  wheels  in  gear. 

27.  If  P  denotes  the  tangential  pressure  at  the  pitch  surface  of 
any  wheel  in  the  train,  and  V  the  velocity  in  feet  per  minute  of  the 
wheel  at  the  pitch  surface,  then, 

PV 

Horse-power  transmitted  =  , 

oo^OOO 

and  this  is  constant  for  any  stage  of  the  transmission. 

Thus,  if  the  lever  handles  make  24  turns  per  minute,  the  horse- 
power at  the  lever  handle  axle  is : 

PV    _  80  X  STT  X  18  X  24  _  n  ,.  ™ 
33^)00  ~          33,000  X  12 

Since  the  value  of  the  train  is  - ,  the  last  wheel  will  make  Q— 


WHEELS  IN  TRAIN  223 

of  a  revolution  in  a  minute,  hence,  the  horse-power  at  the  drum 
axle  is  : 

4827.6  X  2*  X  150  X  3 


33,000  X  12  X  2-  X  10 
In  like  manner,  the  horse-power  transmitted  from  axle  A  to  axle 
J9  may  be  shown  to  be  the  same,  thus  : 

1  8  V  94. 

Axle  A  makes  -  ~  -  =  5.4  revolutions  per  minute,  and  the 
oO 

radius  of  the  pinion  of  40  teeth  is  —  -  —  s  -  =  -^  —  • 

2-  2x 

Horse-power  transmitted  from  axle  A  to  axle  B 

804.6  X  27:  X  50  X  5.4 


33,000  X  2-  X 


28.  The  gearing  of  a  lathe  to  cut  a  thread  of  a  given  pitch  is  a 
familiar  illustration  of  wheels  in  train.  There  are  two  systems  of 
lathe  gearing — simple  and  compound.  The  driving  wheel  of  the 
gearing  is  either  fast  on  the  lathe  spindle,  or  derives  motion  by 
means  of  intermediate  gearing.  The  intermediate  gearing  af- 
fords a  ready  means  of  throwing  the  gear  wheels  out  of  action  when 
the  lathe  is  running  at  high  speeds,  as  for  polishing,  and  it  also 
serves  the  purpose  of  changing  the  direction  of  rotation  of  the 
feed  screw  when  desired.  In  either  case  the  revolutions  of  the  first 
driver  are  the  same  as  those  of  the  lathe  spindle. 

Any  wheels  intervening  between  the  driver  and  the  wheel  on  the 
feed  screw,  no  two  being  on  the  same  axis,  have  no  influence  other 
than  to  convey  the  motion,  and  the  only  wheels  to  be  considered  are 
the  driver  and  the  one  on  the  feed  screw.  If  the  gearing  be  such 
that  the  lathe  spindle  and  the  feed  screw  make  the  same  number  of 


224:    TRANSMISSION  or  POWER  BY  BELTS  AND  GEAR-WHEELS 

turns,  the  thread  cut  would  have  the  same  pitch  as  the  thread  on 
the  feed  screw.  In  all  other  cases  the  pitch  of  the  thread  to  be 
cut  will  be  finer  or  coarser  than  that  on  the  feed  screw  in  the 
exact  proportion  that  the  revolutions  of  the  lathe  spindle  in  a  unit 
of  time  are  greater  or  less  than  those  of  the  feed  screw. 

Figure  10  is  a  representation  of  simple  gearing  for  a  lathe.  The 
wheel  A  is  fast  on  the  lathe  spindle,  or  so  connected  with  it  as  to 
make  the  same  number  of  revolutions  as  the  spindle.  The  wheel 
B  is  on  the  feed  screw  of  the  lathe,  and  the  intermediate  wheel  I  is 
an  idler,  serving  the  purpose  of  making  the  wheels  A  and  B  rotate 
in  the  same  direction,  and  also  of  filling  in  the  space  between 
A  and  B,  the  axis  of  I  being  adjustable  in  a  slotted  arm. 


For  each  revolution  of  the  wheel  B  the  nut  on  the  feed  screw 
advances  a  distance  equal  to  the  pitch  of  the  screw,  and  it  depends 
entirely  upon  the  ratio  between  wheels  B  and  A  as  to  the  number 
of  turns  the  spindle  will  make  while  the  tool  moves  through  the 
same  distance. 

Suppose  the  pitch  of  the  feed  screw  is  ^  inch,  and  it  is  desired  to 
cut  a  thread  of  ^  inch  pitch.  Then  for  two  turns  of  the  feed 
screw  the  tool  will  advance  one  inch,  and  the  lathe  spindle  must  be 

o  o          1  o 

made   to   turn    13    times.      We   have,    then,    e  =  ~  =  -^r  —  ^o  • 

JLo         O/c          <o 

Hence,  giving  12  teeth  to  wheel  A  and  78  to  wheel  B  is  among  the 
possible  solutions. 


WHEELS  IN  TRAIN 


225 


If  it  were  desired  to  cut  10 J  threads  to  the  in^h,  the  feed  screw 
pitch  being  J,  we  would  have  e  =  T?r=  =  ^y  =  ~^-r.  Hence,  giving 

10.0         A\-          o* 

16  teeth  to  the  wheel  A.  and  84  teeth  to  the  wheel  B  is  a  probable 
solution. 

29.  Figure  11  is  a  representation  of  double  gearing  for  a  lathe. 
The  axis  intervening  between  the  lathe  spindle  and  the  feed  screw 
carries  two  wheels  of  different  size  and  are,  therefore,  factors  in 
the  value  of  the  train. 


Suppose  the  pitch  of  the  feed  screw  to  be  -|  inch,  and  it  is  de 
sired   to   cut   20   threads   to   the   inch.     We  would    then   have: 


teeth,  and  40,  24,  and  120  to  B,  C,  and  D  respectively,  is  a  solution. 
In  order  to  cut  a  left-hand  thread  the  lathe  spindle  and  the  feed 
screw  must  turn  in  opposite  directions,  which  may  be  effected  by 
interposing  an  idle  wheel  between  C  and  D. 

30.  The  back  gear  attachment  to  a  lathe  increases  the  power  at 
the  expense  of  the  speed.  Fig.  12  is  a  sketch  illustrating  the  action 
of  back  gear.  The  speed  pulleys,  to  which  the  pinion  A  is  attached 
are  loose  on  the  lathe  spindle.  The  spur-wheel  E  is  keyed  to  the 
spindle.  The  spur-wheel  C  and  pinion  D,  carried  on  the  shaft  F, 
form  the  back  gear  and  can,  at  will,  be  thrown  into  or  out  of  gear 
with  A  and  E. 
15 


226    TRANSMISSION  OF  POWER  BY  BELTS  AND  GEAR-WHEELS 

The  motion  of.  the  speed  pulleys  may  be  conveyed  to  the  lathe 
spindle  in  two  ways :  (a)  The  back  gear  being  disengaged,  as  shown 
in  the  figure,  the  spur-wheel  E  is  made  to  engage  with  the  speed 
cone  by  means  of  a  bolt,  thereby  giving  to  the  spindle  the  same 
revolutions  that  are  made  by  the  cone.  (6)  Disengaging  the  wheel 
E  from  the  speed  cone,  and  throwing  the  back  gear  into  gear  with 
wheels  A  and  E,  the  motion  of  the  speed  pulleys  is  then  transmitted 
to  the  lathe  spindle  by  means  of  the  train  ACDE,  making  the  entire 
system  consist  of  a  train  of  which  the  pulley  on  the  counter-shaft 
is  the  first  driver  and  the  spur-wheel  E  the  last  follower. 

PEOBLEMS. 

21.  Three  spindles,  A,  B,  and  (7,  are  arranged  parallel  to  one 
another.     On  A  there  is  a  spur-wheel  with  52  teeth  which  gears 
with  a  wheel  of  19  teeth  on  spindle  B.     On  spindle  B  is  another 
wheel  of  81  teeth,  gearing  with  one  of  21  teeth  on  spindle  C. 
While  A  is  making  15  turns,  how  many  will  C  make?     How  many 
will  B  make?  Ans.  158.33;   41.05. 

22.  If  the  change  wheels  of  a  lathe  have  18,  30,  40,  50,  and  88 
teeth,  show  an  arrangement  for  cutting  a  screw  of  11  threads  to  the 
inch,  the  lead  screw  having  a  pitch  of  one-third  inch. 

23.  With  a  lead  screw  of  half-inch  pitch,  and  change  wheels  of 
20,  24,  30,  40,  55,  60,  80,  and  100,  show  how  you  would  select 
wheels  to  cut  screws  of  6,  11,  and  16  threads  to  the  inch. 

24.  The  back  gear  of  a  lathe  is  in  use.     Diameter  of  pulley  on 
counter-shaft,  4.25  inches;  diameter  of  pulley  on  lathe,  8.75  inches; 
pinion  on  cone  pulley  of  lathe,  18  teeth;  spur-wheel  on  back  shaft, 
58  teeth;    pinion  on  back  shaft,  18  teeth;    spur  wheel  on  lathe 
spindle,  58  teeth.     How  many  revolutions  per  minute  will  the  lathe 
spindle  make  when  the  counter-shaft  is  making  150  revolutions  per 
minute?  Ans.  7.017. 

25.  Two  parallel  shafts,  whose  axes  are  to  be  as  nearly  as  pos- 
sible 30  inches  apart,  are  to  be  connected  by  a  pair  of  spur  wheels, 
so  that  while  the  driver  runs  at  100  revolutions  per  minute  the  fol- 
lower is  required  to  run  at  only  25  revolutions  per  minute.    Sketch 
the  arrangement,  and  mark  on  each  wheel  its '  pitch-diameter  and 
number  of  teeth,  the  pitch  of  the  teeth  being  1.25  inches.     Deter- 
mine, also,  the  exact  distance  apart  of  the  two  shafts. 

Ans.  29.8295  inches. 


WHEELS  IN  TRAIN  227 

26.  Make  a  sketch  of  a  back  gear  of  a  lathe.    If  the  two  wheels 
have  63  and  63  teeth  each,  and  each  pinion  25  teeth,  find  the  reduc- 
tion in  the  velocity  ratio  of  the  lathe  spindle  due  to  the  back  gear. 

Ans.  I  :  6.35. 

27.  The  double  purchase  wheelwork  of  a  crane  consists  of  a  pin- 
ion of  16  teeth,  on  the  handle  axle;  a  wheel  and  pinion  of  64  and 
20  teeth  respectively,  on  the  first  intermediate  axle;    a  wheel  and 
pinion  of  80  and  18  teeth,  respectively,  on  the  second  intermediate 
axle;  and  a  wheel  of  90  teeth  on  the  drum  axle.     The  power  han- 
dles are  20  inches  long,  the  radius  of  the  drum  11  inches,  the 
diameter  of  the  rope  2  inches,  and  the  pitch  of  the  teeth  1.25  inches. 
Neglecting  friction,  it  is  required  to  find:    (1)    the  power  that 
must  be  applied  at  the  handles  to  raise  9600  Ibs.  at  the  drum;  (2) 
the  tangential  pressures,  P1?  P2,  P3,  between  the  teeth  of  the  wheels 
in  gear;  (3)  the  H.  P.  transmitted,  supposing  the  handles  to  make 
20  turns  per  minute. 

Ans.  72  Ibs;  452.4  Ibs.;  1447.7  Ibs.;  6434  Ibs.;  0.4569  H.  P. 

28.  The  wheelwork  of  a  crane  consists  of  a  pinion  of  11  teeth, 
driven  by  a  lever  handle  18  inches  long,  gearing  with  a  wheel  of  92 
teeth,  and  of  a  pinion  of  12  teeth  gearing  with  a  wheel  of  72  teeth 
on  the  drum  axle.     The  diameter  of  the  drum  being  18  inches,  it  is 
required  to  find  the  ratio  of  the  power  to  the  weight  raised,  friction 
being  neglected.  Ans.  1  :  100  nearly. 

29.  The  motion  of  an  engine  shaft  is  communicated  to  a  shaft, 
A}-  by  means  of  an  open  belt  passing  over  a  fly-wheel  64  inches  in 
diameter  to  a  15-inch  pulley.     The  motion  of  shaft  A  is  trans- 
mitted to  a  shaft,  B,  by  means  of  spur-wheels  of  J-inch  pitch  and 
of  72  and  24  teeth  respectively,  with  an  intermediate  wheel.     The 
motion  of  shaft  B  is  communicated  to  the  spindle  of  a  fan  by 
means  of  an  open  belt  passing  over  pulleys  of  24  and  12  inches 
diameter  respectively.     The  revolutions  of  the  engine  are  200  per 
minute,  the  stroke  16  inches,  and  the  mean  pressure  on  the  crank 
pin  8000  pounds.     Find  the  revolutions  per  minute  of  the  fan,  and 
the  driving  tensions  in  the  belts  passing  over  the  pulleys  on  the 
shafts  A  and  B.     Will  the  fan  and  fly-wheel  turn  in  the  same  or  in 
opposite  directions? 

Ans.  5120 -revolutions;  2000  Ibs.;  416.66  Ibs. 


PART  III 
MATERIALS  OF  ENGINEERING 


CHAPTEE  I. 
MATERIALS. 

1.  COPPER. — In  the  pure  state  copper  is  red  in  color,  soft,  ductile, 
and  malleable,  with  a  melting  point  at  about  2000°  F.,  and  a  tensile 
strength  of  from  20,000  Ibs.  to  30,000  Ibs.  per  square  inch  of  sec- 
tion.    It  is  not  readily  welded,  except  electrically,  but  is  easily 
joined  by  the  operation  of  brazing.     Attempts  have  been  made  to 
temper  it,  but  without  much  practical  success.    It  is  readily  forged 
and  cast,  and,  when  cold,  may  be  rolled  into  sheets  or  drawn  into 
wire.    When  in  sheets  or  in  small  pieces  it  may  be  spun,  flanged, 
and  worked  under  the  hammer. 

The  tensile  strength  of  copper  rapidly  falls  off  as  the  temperature 
rises  above  about  400°  F.,  so  that  from  800°  to  900°  the  strength 
is  only  about  one-half  what  it  is  at  ordinary  temperatures.  This 
peculiarity  of  copper  should  be  borne  in  mind  when  it  is  used  in 
places  where  the  temperature  is  liable  to  rise  to  these  figures. 
Again,  if  copper  is  raised  nearly  to  its  melting  point  in  contact 
with  air,  it  readily  unites  with  oxygen  and  loses  its  strength  in 
large  degree,  becoming,  when  cool,  crumbly  and  brittle.  Copper 
in  this  condition  is  said  to  have  been  burned.  The  possibility  of 
thus  injuring  the  tenacity  of  copper  is  of  the  highest  importance  in 
connection  with  the  use  of  brazed  joints  in  steam  pipes. 

2.  In  the  'operation  of  brazing  a  joint,  the  surfaces  to  be  joined 
are  cleaned,  bound  together  with  wire  or  otherwise,  then  supplied 
with  brazing  solder  in  small  bits,  mixed  with  borax  as  a  flux,  and 
placed  in  a  clear  fire  until  the  solder  melts  and  forms  the  joint. 
The  brazing  solder,  or  hard  solder,  as  it  is  often  called,  is  usually 
a  brass  or  alloy  of  copper  and  zinc.     The  melting  point  of  all  such 
alloys  is  below  that  of  copper,  and  when  copper  is  joined  to  brass,  or 
two  pieces  of  brass  are  joined  together,  the  solder  used  must  have 
a  melting  point  lower  than  either  of  these  metals.     In  the  operation 
of  brazing  a  copper  joint,  therefore,  the  greatest  care  must  be  taken 
in  the  selection  of  a  solder  and  in  attention  to  the  fire,  so  that  there 
may  be  no  danger  of  burning  the  copper,  and  thus  endangering  the 
quality  of  the  metal  in  the  joint. 


232  MATERIALS  OF  ENGINEERING 

Copper  unalloyed  is  used  chiefly  for  pipes  and  fittings,  especially 
for  junctions,  elbows,  bends,  etc.  For  large  sizes  the  material  is 
made  in  sheets,  bent  and  formed  to  the  desired  shape  and  brazed  at 
the  seams.  Small  sizes  are  either  made  by  the  same  general  process 
or  drawn  from  the  solid,  which  then  may  be  bent  as  desired  after 
drawing.  Copper  is  also  largely  used  as  the  chief  ingredient  of  the 
various  brasses  and  bronzes. 

3.  CAST  IRON. — This  material  consists  of  a  mixture  and  combina- 
tion of  iron  and  carbon,  with  other  substances  in  varying  propor- 
tions. 

Influence  of  Carbon. — In  the  molten  condition,  the  carbon  is 
dissolved  by  the  iron  and  held  in  solution  just  as  ordinary  salt  is 
dissolved  by  water.  The  mixture  or  combination  of  the  two  ele- 
ments is  thus  entirely  uniform.  The  proportion  of  carbon  which 
pure  melted  iron  can  thus  dissolve  and  hold  in  solution  is  about 
3.5  per  cent.  If  chromium  or  manganese  is  present  also,  the 
capacity  for  carbon  is  much  increased,  while  with  silicon,  on  the 
other  hand,  the  capacity  for  carbon  is  decreased.  In  the  various 
grades  of  cast  iron,  the  proportion  of  carbon  is  usually  found  to  be 
between  2  per  cent  and  4.5  per  cent. 

Now,  when  such  a  molten  mixture  cools  and  becomes  solid,  there 
is  a  tendency  for  a  part  of  the  carbon  to  be  separated  and  no  longer 
remain  in  intimate  combination  with  the  iron.  The  carbon  thus 
separated,  or  precipitated,  from  the  iron  takes  that  form  known  as 
graphite,  and  collects  in  very  small  flakes  or  scales.  The  carbon 
which  remains  in  intimate  combination  with  the  iron  is  said  to  be 
combined,  while  that  which  is  separated  is  usually  called  graphitic. 

The  qualities  of  cast  iron  depend  chiefly  on  the  proportion  of 
total  carbon  and  on  the  relative  proportions  of  combined  and 
graphitic  carbon. 

With  a  high  proportion  of  graphitic  carbon  the  iron  is  soft  and 
tough,  with  a  low  tensile  strength,  and  breaks  with  a  dark  and 
coarse-grained  fracture.  In  fact,  the  substance  in  this  condition 
may  be  considered  as  nearly  pure  iron  with  fine  flakes  of  graphite 
entangled  and  distributed  through  it,  thus  giving  to  the  iron  a 
spongy  structure.  The  iron  thus  forms  a  kind  of  continuous  mesh 
'  about  the  graphite,  which  decreases  the  strength  by  reason  of  the 
decrease  of  cross-sectional  area  actually  occupied  by  the  iron  itself. 
Such  irons  are  termed  gray. 


MATERIALS  233 

As  the  relative  proportion  of  graphitic  carbon  decreases  and  that 
of  combined  carbon  increases,  the  iron  takes  on  new  properties,  be- 
coming harder  and  more  brittle.  Its  tensile  strength  also  increases 
to  a  certain  extent,  and  the  fracture  becomes  fine-grained,  or  smooth, 
and  whiter  in  color.  When  these  characteristics  are  pronounced, 
the  iron  is  said  to  be  white.  When  about  half  the  carbon  is  com- 
bined and  half  separates  as  graphite,  the  effect  is  to  produce  a  dis- 
tribution of  dark  spots  scattered  over  a  whitish  field.  Such  iron  is 
said  to  be  mottled. 

In  a  general  way,  with  a  large  proportion  of  total  carbon,  there 
is  likely  to  be  formed  a  considerable  amount  of  graphitic  carbon, 
and  hence  such  iron  is  usually  gray  and  soft.  With  a  large  propor- 
tion of  carbon  also,  the  iron  melts  more  readily. and  its  fluidity  is 
more  pronounced.  As  the  proportion  of  total  carbon  decreases,  the 
cast  iron  approaches  gradually  the  condition  of  steel,  whose  prop- 
erties will  be  discussed  in  later  paragraphs. 

Of  the  special  ingredients  in  cast  iron  the  combined  carbon  is 
one  of  greatest  importance.  It  is  that  chiefly  which,  by  uniting 
with  the  iron,  gives  it  new  qualities,  and  the  principal  influence  of 
other  substances  lies  in  the  effect  which  they  may  have  on  the  pro- 
portion of  this  ingredient.  As  between  graphitic  and  combined 
carbon,  the  former  does  not  affect  the  quality  of  the  iron  itself,  but 
acts  physically  by  affecting  the  structure  of  the  casting,  while  the 
latter  by  entering  into  combination  with  the  iron,  acts  chemically, 
and  produces  a  new  substance  with  different  qualities. 

The  proportions  of  combined  and  graphitic  carbon  are  influenced 
by  the  rate  of  cooling,  and  by  the  presence  or  absence  of  various 
other  ingredients.  Slow  cooling  allows  time  for  the  separation  of 
the  carbon  and  thus  tends  to  form  graphitic  carbon  and  soft  gray 
irons.  Quick  cooling,  or  chilling  in  the  extreme  case,  prevents  the 
formation  of  graphitic  carbon  and  thus  tends  to  form  hard,  white 
iron. 

In  addition  to  carbon,  small  particles  of  silicon,  sulphur,  phos- 
phorus, manganese,  and  chromium  may  be  found  in  cast  iron. 

4.  Influence  of  Silicon. — The  fundamental  influences  of  silicon 
are  two.  (a)  It  tends  to  expel  the  carbon  from  the  combined  state 
and  thus  to  decrease  the  relative  proportion  of  combined  carbon  and 
increase  that  of  graphitic  carbon.  (6)  Of  itself  silicon  tends  to 
harden  cast  iron  and  to  make  it  brittle. 


234  MATERIALS  OF  ENGINEERING 

These  two  influences  are  opposite  in  character,  since  an  increase 
in  graphitic  carbon  softens  the  iron.  In  usual  cases  the  net  result 
is  a  softening  of  the  iron,  an  increase  in  fluidity,  and  a  general 
change  toward  those  qualities  possessed  by  iron  with  a  high  pro- 
portion of  graphitic  carbon.  This  applies  with  a  proportion  of 
silicon  from  2  to  4  per  cent.  With  more  than  this  the  influence  on 
the  carbon  is  but  slight  and  the  result  on  the  iron  is  to  decrease  the 
strength  and  toughness,  giving  a  hard  but  brittle  and  weak  grade 
of  iron. 

A  chilled  cast  iron  is  an  iron  which  if  cooled  slowly  would  be 
gray  and  soft,  but  if  cooled  suddenly  by  contact  with  a  metal  mould 
or  other  means,  becomes  white  or  hard,  especially  at  and  near  the 
surfaces.  Certain  grades  of  cast  iron  tend  to  chill  when  cast  in 
sand  moulds.  This  property  is  usually  undesirable.  In  such  cases 
the  tendency  can  be  prevented  by  the  addition  of  silicon,  which,  by 
forcing  the  carbon  into  the  graphitic  state  on  cooling,  prevents  the 
formation  of  hard,  chilled  surfaces.  In  all  cases  the  actual  effect 
of  adding  silicon  will  depend  much  on  the  character  of  the  iron 
used  as  a  base,  and  only  a  statement  of  the  general  tendencies  can 
here  be  given. 

To  sum  up,  a  white  iron  which  would  give  hard,  brittle,  and  por- 
ous castings  can  be  made  softer,  tougher,  and  more  solid  by  the  addi- 
tion of  silicon  to  the  extent  of  perhaps  2  or  3  per  cent.  As  the 
silicon  is  increased,  the  iron  will  become  softer  and  grayer  and  the 
tensile  strength  will  decrease.  At  the  same  time  the  shrinkage  will 
decrease,  at  least  for  a  time,  though  it  may  increase  again  with 
large  excess  of  silicon.  The  softening  and  toughening  influence, 
however,  will  only  continue  so  long  as  additional  graphite  is  formed, 
and  when  most  of  the  carbon  is  brought  into  this  state,  the  maxi- 
mum effect  has  been  produced  and  any  further  addition  of  silicon 
will  decrease  both  strength  and  toughness. 

5.  Influence  of  Sulphur. — Authorities  are  not  in  entire  agree- 
ment as  to  the  influence  of  sulphur  on  cast  iron,  some  believing 
that  it  tends  to  increase  the  proportion  of  combined  carbon,  while 
others  maintain  that  it  tends  to  decrease  both  the  combined  carbon 
and  silicon.  It  is  generally  agreed,  however,  that  in  proportions 
greater  than  about  0.15  to  0.20  of  1  per  cent,  it  increases  the  shrink- 
age and  tendency  to  chill,  and  decreases  the  strength.  Sulphur 
does  not,  however,  readily  enter  cast  iron  under  ordinary  condi- 
tions, and  its  influence  is  not  especially  feared.  An  increase  in  the 


MATERIALS  235 

proportion  of  sulphur  in  cast  iron  is  most  likely  to  result  from  an 
absorption  of  sulphur  in  the  coke  during  the  operation  of  melting 
in  the  cupola. 

6.  Influence   of  Manganese. — This   element  by  itself   decreases 
fluidity,  increases  shrinkage,  and  makes  the  iron  harder  and  more 
brittle.     It  combines  with  iron  in  all  proportions.     The  combina- 
tion containing  less  than  50  per  cent  of  manganese  is  called  spiegel- 
eisen.     With  more  than  50  per  cent  of  manganese,  it  is  called 
ferromanganese.     One  of  the  most  important  properties  of  man- 
ganese in  combination  with  iron  is  that  it  increases  the  capacity  of 
the  iron  for  carbon.     Pure  iron  will  only  take  about  3.5  per  cent 
of  carbon,  while  with  the  addition  of  manganese  the  proportion 
may  rise  to  6  or  7  per  cent.     Manganese  is  also  believed  to  decrease 
the  capacity  of  iron  for  sulphur,  and  to  this  extent  may  be  a  de- 
sirable ingredient  in  proportions  not  exceeding  1  to  1.5  per  cent. 

7.  Influence  of  Chromium. — This  substance  is  rarely  found  in 
cast  iron,  but  it  has  the  property,  when  present  in  large  propor- 
tion, of  raising  the  capacity  of  the  iron  for  carbon  from  3.5  per 
cent  up  to  about  12  per  cent. 

8.  Shrinkage  of  Cast  Iron. — At  the  moment  of  hardening,  cast 
iron  expands  and  takes  a  good  impression  of  the  mould.     In  the 
gradual  cooling  after  setting,  however,  the  metal  contracts,  so  that 
en  the  whole  there  is  a  shrinkage  of  about  -J  inch  per  foot  in  all 
directions,  though  this  amount  varies  somewhat  with  the  quality 
of  the  iron  and  with  the  form  and  dimensions  of  the  pattern.     In 
a  general  way  hardness  and  shrinkage  increase  and  decrease  to- 
gether. 

9.  Strength  and  Hardness  of  Cast  Iron. — The  hardness  of  cast 
iron  is  chiefly  dependent  on  the  amount  of  combined  carbon,  as 
already  noted. 

The  strength  is  also  chiefly  dependent  on  the  same  ingredient. 
The  greatest  crushing  strength  is  obtained  with  sufficient  combined 
carbon  to  make  a  rather  hard,  white  iron,  while  for  the  maximum 
transverse  or  bending  strength,  the  combined  carbon  is  somewhat 
less  and  the  iron  only  moderately  hard;  and  for  the  greatest  ten- 
sile strength,  the  combined  carbon  is  still  less  and  the  iron  rather 
soft.  Metal  still  softer  than  this  works  with  the  greatest  facility, 
but  is  deficient  in  strength. 


236  MATERIALS  OF  ENGINEERING 

10.  Use  of  Cast  Iron  .in  Engineering. — Cast  iron  is  used  for 
cylinders,  cylinder  heads,  liners,  slide  valves,  valve  chests  and  con- 
nections, and  generally  for  all  parts  having  considerable  complexity 
of  form.     It  is  also  used  for  columns,  bed  plates,  bearing  pedestals, 
caps,  etc.,  though  cast  and  forged  steel  are  to  some  extent  displac- 
ing cast  iron  for  some  of  these  items.     It  is  also  used  for  grate 
bars,  furnace  door  frames,  and  minor  boiler  fittings,  and  for  a 
great  variety  of  special  purposes  usually  connected  with  the  sta- 
tionary or  supporting  parts  of  machines. 

11.  Inspection  of  Castings. — In  the  inspection  of  castings,  care 
must  be  had  to  note  the  texture  of  the  surface,  and  to  this  end  the 
outer  scale  and  burnt  sand  should  be  carefully  removed  by  the  use 
of  brushes  or  chipping  hammer,  or,  if  necessary,  by  pickling  in 
dilute  muriatic  acid.     The  flaws  most  liable  to  occur  are  blow 
holes  and  shrinkage  cracks.     The  latter,  however,  are  not  often 
met  with  when  the  moulding  and  casting  are  properly  carried  out. 
The  parts  of  the  casting  most  liable  to  be  affected  by  blow  holes 
are  those  on  the  upper  side  or  near  the  top.     On  this  account  a 
sinking  head  or  extra  piece  is  often  cast  on  top,  into  which  the 
gases  and  impurities  may  collect.     This  is  afterward  cut  off,  leav- 
ing the  sounder  metal  below. 

The  presence  of  blow  holes,  if  large  in  size  or  in  great  number 
and  near  the  surface,  may  often  be  determined  by  tapping  with  a 
hand  hammer.  The  sound  given  out  will  serve  to  indicate  to  an 
experienced  ear  the  probable  character  of  the  metal  underneath. 

12.  Cast  iron  may  be  brazed  to  itself  or  to  most  of  the  structural 
metals  by  the  use  of  a  brazing  solder  of  suitable  melting  point,  and 
with  proper  care  in  the  operation.     Cast  iron  may  also  be  united 
to  itself  or  to  wrought  iron  or  steel  by  the  operation  of  burning. 
This  consists  in  placing  in  position  the  two  pieces  to  be  united,  and 
then  allowing  a  stream  of  melted  cast  iron  to  flow  over  the  surfaces 
to  be  joined,  the  adjacent  parts  being  protected  by  fire  clay  or 
other  suitable  material.     The  result  is  to  soften  or  partially  melt 
the  surfaces  of  the  pieces,  and  by  arresting  the  operation  at  the 
right  moment  they  may  be  securely  joined  together. 

13.  Malleable  Cast  Iron. — If  iron  castings  of  not  too  great  thick- 
ness, and  of  such  purity  as  to  be  low  in  sulphur,  be  embedded  in 
powdered  red  oxide  of  iron  (red  hematite)   and  maintained  at  a 
red  heat  for  two  or  three  days,  they  become,  in  a  measure,  decar- 


MATERIALS  237 

bonized  as  a  result  of  the  chemical  action  which  ensues.  The  car- 
bon first  disappears  from  the  outer  layers,  and  as  the  process  con- 
tinues, decarbonization  toward  the  innermost  layers  takes  place. 
If  the  process  be  carried  to  the  extreme  in  the  effort  to  withdraw 
all  the  carbon  from  the  interior,  the  outer  layer  is  very  liable  to 
become  brittle,  thus  defeating  the  object  to  be  attained.  For  this 
reason  there  always  remains  a  core  of  cast  iron  only  partially  de- 
carbonized, while  the  outermost  layers  are  left  in  the  condition  of 
soft,  or  malleable,  iron.  The  process  has  little  effect  upon  the 
sulphur,  manganese,  phosphorus,  and  other  impurities  of  the  cast- 
ings, but  the  resulting  product  is  a  malleable  casting  much  less 
fusible  than  cast  iron  and  possessing  six  times  its  ductility.  The 
best  product  can  be  twisted  and  bent  to  a  considerable  extent  before 
breaking,  and  its  ability  to  withstand  shocks  is  much  greater  than 
that  of  cast  iron.  Pipe  fittings,  to  some  extent,  are  malleable 
castings,  and  this  material  is  largely  used  in  appliances  of  car 
construction  where  more  strength  and  toughness  are  required  than 
cast  iron  affords. 

14.  WROUGHT  IRON. — Wrought  iron  is  nearly  pure  iron  mixed 
with  more  or  less  slag.  Nearly  all  the  wrought  iron  used  in  mod- 
ern times  is  made  by  the  puddling  process.  For  the  details  of  this 
process  reference  may  be  had  to  text-books  on  metallurgy.  We  can 
only  note  here  that,  in  a  furnace  somewhat  similar  to  the  open- 
hearth,  most  of  the  carbon,  silicon,  and  other  special  ingredients  of 
cast  iron  are  removed  by  the  combined  action  of  the  flame  and  of 
a  molten  bath  of  slag  or  fluxing  material,  consisting  chiefly  of 
black  oxide  of  iron.  As  this  process  approaches  completion  small 
bits  of  nearly  pure  iron  separate  from  the  bath  of  melted  slag  and 
unite  together.  This  operation  is  assisted  by  the  puddling  bar, 
and  after  the  iron  has  thus  become  separated  from  the  liquid  slag 
it  is  taken  out,  hammered  or  squeezed,  and  rolled  into  bars  or  plates. 
Some  of  the  slag  is  necessarily  retained  in  the  iron  and,  by  the 
process  of  manufacture,  is  drawn  out  into  fine  threads,  giving  to  the 
iron  a  stringy  or  fibrous  appearance  when  nicked  and  bent  over 
or  when  pulled  apart. 

The  proportion  of  carbon  in  wrought  iron  is  very  small,  ranging 
from  0.02  to  0.20  of  1  per  cent.  In  addition,  small  amounts  of 
sulphur,  phosphorus,  silicon,  and  manganese  are  usually  present. 

The  proportion  of  sulphur  should  not  exceed  0.01  of  1  per  cent. 


238  MATERIALS  OF  ENGINEERING 

Excess  of  sulphur  makes  the  iron  red-short,  that  is,  brittle  when 
red  hot. 

The  proportion  of  phosphorus  may  vary  from  0.05  to  0.25  of 
1  per  cent.  Excess  of  phosphorus  makes  the  metal  cold-short, 
that  is,  brittle  when  cold. 

The  proportion  of  silicon  may  vary  from  0.05  to  0.30  of  1  per 
cent. 

The  proportion  of  manganese  may  vary  from  0.005  to  0.050  of 
1  per  cent.  The  influence  of  the  silicon  and  manganese  is  usually 
slight  and  unimportant. 

Special  Properties. — Wrought  iron  is  malleable  and  ductile,  and 
may  be  rolled,  flanged,  forged,  and  welded.  It  cannot  be  hard- 
ened as  steel,  though  by  the  process  of  case-hardening  a  surface 
layer  of  steel  is  formed  and  may  be  hardened.  Wrought  iron  may 
be  welded,  because  for  a  considerable  range  of  temperature  below 
melting  (which  takes  place  only  at  a  very  high  temperature)  the 
iron  becomes  soft  and  plastic,  and  two  pieces  pressed  together  in 
this  condition  unite  and  form,  on  cooling,  a  junction  nearly  as 
strong  as  the  solid  metal.  In  order  that  this  welding  operation  be 
successful,  the  iron  must  be  heated  sufficiently  to  bring  it  to  the 
plastic  condition,  yet  not  overheated,  and  there  must  be  employed 
a  flux  (usually  borax)  which  will  unite  with  the  iron  oxide  and 
other  impurities  at  the  joint,  and  form  a  thin  liquid  slag,  which 
may  readily  be  pressed  out  in  the  operation,  thus  allowing  the 
clean  metal  surfaces  of  the  iron  to  effect  a  union  as  desired. 

15.  Uses  in  Marine  Engineering. — In  modern  practice,  the  place 
of  wrought  iron  in  marine  engineering  has  almost  entirely  been 
taken  by  steel.     Its  former  office  was  for  all  moving  parts  requir- 
ing strength  and  toughness.     It  is  still  used  to  some  extent  for  stay 
bolts  and  braces  of  boilers,  and  for  boiler  tubes. 

16.  STEEL. — The  properties  of  steel  depend  partly  on  the  propor- 
tions of  carbon  and  other  ingredients  which  it  may  contain,  and 
partly  on  the  process  of  manufacture.     The  proportion  of  carbon 
is  intermediate  between  that  for  wrought  iron  and  for  cast  iron. 
In  the  so-called  mild  or  structural  steel,  the  carbon  is  usually  from 
3^-  to  -J  of  1  per  cent.     In  spring  steel  the  carbon  proportion  is 
somewhat  greater,  and  in  high  carbon  grades,  such  as  are  used  for 
tool  steel,  etc.,  the  carbon  is  from  0.6  to  1.2  per  cent.     In  addition 


MATERIALS  239 

to  the  carbon  there  may  be  sulphur,  phosphorus,  silicon,  and  man- 
ganese in  varying  but  very  small  amounts. 

Steel  may  be  made  from  wrought  iron  by  increasing  its  propor- 
tion of  carbon,  or  from  cast  iron  by  decreasing  the  proportion  of 
carbon.  The  earlier  processes  followed  the  first  method,  and  high- 
grade  steels  are  still  made  in  this  way  by  the  crucible  process. 

17.  Crucible  Steel. — In  this  process  a  pure  grade  of  wrought 
iron  is  rolled  out  into  flat  bars.     These  are  then  cut,  and  piled 
and  packed  with  intermediate  layers  of  charcoal,  and  subjected  to 
a  high  temperature  for  several  days.     This  recarbonizes  or  adds 
carbon  to  the  wrought  iron,  and  thus  makes  what  is  then  called 
cement  or  blister  steel.     These  bars  are  then  broken  into  pieces  of 
convenient  size,  placed  in  small  crucibles,  melted,  and  cast  into 
bars  or  into  such  shapes  as  are  desired. 

18.  Mild,  or  Structural  Steel,  is  made  wholly  by  the  second  gen- 
eral method — the  reduction  of  the  proportion  of  carbon  in  cast 
iron.     There  are  two  general  processes,  known  as  the  Bessemer, 
and  the  Siemens-Martin  or  open-hearth. 

19.  Bessemer  Process. — In  this  process  the  carbon  and  silicon 
are  burned  almost  entirely  out  of  the  cast  iron  by  forcing  an  air 
blast  through  the  molten  iron  in  a  vessel  known  as  a  converter.     A 
small  amount  of  spiegeleisen,  or  iron  rich  in  carbon  and  man- 
ganese, is  then  added  in  such  weight  as  to  make  the  proportion  df 
carbon  and  manganese  suitable  for  the  charge  as  a  whole.     The 
steel  thus  formed  is  then  cast  into  ingots  or  into  such  forms  as 
may  be  desired.     In  this  process  no  sulphur  or  phosphorus  is  re- 
moved, so  that  it  is  necessary  to  use  a  cast  iron  nearly  free  from 
these  ingredients  in  order  that  the  steel  may  have  the  properties 
desired. 

A  modification  by  means  of  which  the  phosphorus  is  removed, 
and  known  as  the  basic  Bessemer  process,  is  used  to  some  extent. 
In  this,  calcined  or  burnt  lime  is  added  to  the  charge  just  before 
pouring.  This  unites  with  the  phosphorus,  removes  it  from  the 
steel,  and  brings  it  into  the  slag.  In  the  basic  process,  the  lining 
of  the  converter  is  made  of  ganister  or  a  calcined  magnesia  lime- 
stone, in  order  that  it  may  not  also  be  attacked  by  the  added  lime- 
stone and  the  resulting  slag. 


240  MATERIALS  OF  ENGINEERING 

In  the  Bessemer  process  first  noted,  and  often  known  as  the  acid 
process,  in  distinction  from  the  basic  process,  the  lining  of  the 
converter  is  of  ordinary  fire  clay  or  like  material. 

The  removal  of  the  phosphorus  by  the  basic  process  makes  pos- 
sible the  use  of  an  inferior  grade  of  cast  iron.  At  the  same  time, 
engineers  are  not  altogether  agreed  as  to  the  relative  values  of  the 
two  products,  and  many  prefer  steel  made  by  the  acid  process  from 
an  iron  nearly  free  from  phosphorus  at  the  start. 

20.  The  Open-hearth  Process. — In  this  process  a  charge  of  ma- 
terial consisting  of  wrought  iron,  cast  iron,  steel  scrap,  and  some- 
times certain  ores,  is  melted  on  the  hearth  of  a  reverberatory  fur- 
nace heated  by  gas  fuel  on  the  Siemens-Martin,  or  regenerative 
system.  The  carbon  is  thus  partially  burned  out  in  much  the 
same  manner  as  for  wrought  iron,  and  the  proportion  of  carbon  is 
brought  down  to  the  desired  point  or  slightly  below.  A  charge  of 
spiegeleisen  or  of  ferromanganese  is  then  added  in  order  that  the 
manganese  may  act  on  any  oxide  of  iron  slag  which  remains  in  the 
bath,  and  which  would  make  the  steel  red-short  if  allowed  to  form 
a  part  of  the  charge.  The  manganese  separates  the  iron  from  the 
oxide,  returns  it  to  the  bath,  while  the  carbon  joins  with  that  already 
present,  and  thus  produces  the  desired  proportions. 

Here,  as  with  the  similar  operation  with  the  Bessemer  con- 
verter, there  is  no  removal  either  of  sulphur  or  of  phosphorus,  and 
only  materials  nearly  free  from  these  ingredients  can  be  used  for 
steel  of  satisfactory  quality.  With  very  low  carbon,  however,  a 
little  phosphorus  seems  to  be  desirable  to  add  strength  to  the  metal. 
This  limitation  of  the  available  materials  has  led,  as  with  the 
Bessemer  process,  to  the  use  of  calcined  limestone,  which  unites 
with  most  of  the  phosphorus  and  holds  it  in  the  slag.  Here,  as 
in  the  Bessemer  process  also,  it  is  necessary  to  use  a  basic  lining 
for  the  furnace,  and  it  is  known  as  the  basic  open-hearth  process. 
The  method  without  the  use  of  the  limestone  has  come  to  be  known 
as  the  acid  open-hearth  process. 

As  between  the  products  of  the  two  open-hearth  processes,  there 
is  much  difference  of  opinion  among  engineers.  Either  will  pro- 
duce good  steel  with  proper  care,  and  neither  will  without  it.  It 
is  usually  considered  sufficient  to  specify  the  allowable  limits  for 
the  proportions  of  phosphorus  and  sulphur,  and  leave  the  choice  of 
the  acid  or  basic  process  to  the  maker. 


MATERIALS  241 

21.  Open-hearth  and  Bessemer  Steels  Compared. — Open-hearth 
steel  is  usually  preferred  for  structural  material  in  marine  engi- 
neering.    This  is  because: 

(a)  It  seems  to  be  more  reliable  and  less  subject  to  unexpected 
or  unexplained  failure  than  the  Bessemer  product. 

(&)  Analysis  shows  that  it  is  much  more  homogeneous  in  com- 
position than  Bessemer  steel,  and  experience  shows  that  it  is  much 
more  uniform  in  physical  quality.  This  is  due  to  the  process  of 
manufacture,  which  is  much  more  favorable  to  a  thorough  mixing 
of  the  charge  than  in  the  Bessemer  process. 

(c)  The  open-hearth  steel  may  be  tested  from  time  to  time  dur- 
ing the  operation,  so  that  its  composition  may  be  determined  and 
adjusted  to  fulfil  specified  conditions.  This  is  not  possible  with 
the  Bessemer  process. 

22.  Influence  of  Sulphur  on  Steel. — Sulphur  makes  steel  red- 
short  or  brittle  when  hot,  and  interferes  with  its  forging  and  weld- 
ing properties.     Manganese  tends  to  counteract  the  bad  effects  of 
sulphur.     Good  crucible  steel  has  rarely  more  than  0.01  of  1  per 
cent  of  sulphur.     In  structural  steel  the  proportion  may  vary  from 
0.02  to  0.08  or  0.10  of  1  per  cent.     When  possible  it  should  be 
reduced  to  not  more  than  0.03  or  0.04  of  1  per  cent. 

23.  Influence  of  Phosphorus  on  Steel. — Phosphorus  increases  the 
tensile  strength  and  raises  the  elastic  limit  of  low  carbon  or  struc- 
tural steel,  but  at  the  expense  of  its  ductility  and  toughness,  or 
ability  to  withstand  shocks  and  irregularly  applied  loads.     It  is  thus 
considered  as  a  dangerous  ingredient,  and  the  amount  allowable 
should  be  carefully  specified.     This  is  usually  placed  from  0.02  to 
0.10  of  1  per  cent. 

24.  Influence  of  Silicon  on  Steel. — Silicon  tends  to  increase  the 
tensile  strength  and  reduce  the  ductility  of  steel.     It  also  increases 
the  soundness  of  castings  and  ingots,  and  by  reducing  the  iron 
oxide,  tends  to  prevent  red-shortness.     The  process  of  manufacture 
usually  removes  nearly  all  of  the  silicon,  so  that  it  is  not  an  element 
likely  to  give  trouble  to  the  steel  makers.     The  proportion  allow- 
able should  not  be  more  than  from  0.10  to  0.20  of  1  per  cent. 

25.  Influence  of  Manganese  on  Steel. — This  element  is  believed 
to  increase  hardness  and  fluidity,  and  to  raise  the  elastic  limit  and 
increase  the  tensile  strength.     It  also  removes  iron  oxide  and  sul-         * 

16 


242  MATERIALS  OF  ENGINEERING 

phur,  and  tends  to  counteract  the  influence  of  such  amounts  of 
sulphur  and  phosphorus  as  may  remain.  It  is  thus  an  important 
factor  in  preventing  red-shortness.  The  proportion  needed  to  ob- 
tain these  valuable  effects  is  usually  found  between  0.20  and  0.50 
of  1  per  cent. 

26.  Semi-steel. — A  metal  bearing  this  trade  name  has  in  recent 
years  attracted  favorable  attention  among  engineers  and  has  come 
into  considerable  use  where  somewhat  greater  strength  and  tough- 
ness are  required  than  can  be  provided  by  cast  iron.     It  is  made  by 
melting  mild  steel  scrap,  such  as  punchings  and  clippings  of  boiler 
plate,  with  cast  iron  pig,  in  the  proportion  of  25  or  30  per  cent 
of  the  former  to  75  or  70  per  cent  of  the  latter.     The  presence  of 
manganese  and  other  special  fluxes  in  small  proportions  is  also 
found  to  add  essentially  to  the  strength,  toughness,  and  ability  to 
withstand  shocks  decidedly  greater  than  for  cast  iron,  and  with 
fairly  good  machine  qualities.     Semi-steel  casts  as  readily  as  most 
grades  of  cast  iron,  and  its  shrinkage  and  general  manipulation  are 
about  the  same.     The  chief  drawback  seems  to  lie  in  the  danger  of 
hardness  under  the  lathe,  planer,   or  boring  tool,  but  with  the 
proper  mixture  this  is  avoided,  and  a  material  very  satisfactory  for 
many  purposes  in  marine  engineering  is  thus  produced. 

27.  Mechanical  Properties   of  Steel. — The  tensile  strength  of 
the  lowest  carbon  steel,  say  about  0.10  of  1  per  cent  carbon,  is 
usually  not  above  from  50,000  to  55,000  Ibs.  per  square  inch  of 
section.     The  strength  increases  with  the  increase  of  carbon  and, 
with  not  above  the  usual  proportions  of  sulphur  and  phosphorus, 
quite   uniformly.     Experiment   shows   that   under   these    circum- 
stances the  strength  will  increase  up  to  75,000  Ibs.  per  square  inch, 
or  higher,  at  the  rate  of  from  1200  to  1500  Ibs.  per  0.01  of  1  per 
cent  of  carbon  added.     At  the  same  time,  with  the  increase  in 
strength  the  ductility  decreases,  so  that  a  proper  choice  must  be 
made  according  to  the  particular  uses  for  which  the  steel  is  intended. 
In  the  best  grades  of  tool  steel,  with  carbon  ranging  from  0.5  to 
1.0  per  cent  or  over,  the  strength  ranges  from  80,000  to  120,000  Ibs., 
and  even  higher  in  exceptional  cases. 

28.  Flange  and  Rivet  Steel  must  be  tough  and  ductile  in  the 
highest  degree.     Such  steel  has  usually  a  tensile  strength  between 
50,000  and  60,000  Ibs.  and  an  elastic  limit  of  30,000  to  40,000  Ibs. 
Its  elongation  in  8  inches  is  from  30  to  35  per  cent,  and  the  reduc- 


MATERIALS  243 

tion  of  area  at  the  ruptured  section  from  50  to  60  per  cent.  It 
will  bend  cold  on  itself  and  close  down  flat  under  hammer  or  press 
without  a  sign  of  fracture,  the  thickness  not  exceeding  1  inch. 

29.  Shell  Steel,  used  for  boiler  shells,  etc.,  has  usually  a  strength 
between   55,000   and   65,000   Ibs.;    elastic   limit   from   33,000   to 
45,000  Ibs.;  elongation  in  8  inches  of  25  to  30  per  cent,  and  reduc- 
tion of  area  at  the  fractured  section  of  from  50  to  60  per  cent. 

30.  For  Shafting,  the  quality  of  the  steel  is  about  the  same  as 
for  shell  plates.     For  piston  and  connecting-rods,  the  strength  is 
rather  higher,  but  ductility  somewhat  lower. 

31.  For  Steel  Castings,  the  strength  required  is  usually  from 
60,000  to  65,000  Ibs.,  with  an  elongation  in  8  inches  of  from  10 
to  15  per  cent. 

32.  Special  Properties  of  Steel. — Mild  or  low  carbon  steel  may 
be  welded,  forged,  flanged,  rolled,  and  cast.     It  cannot  be  tem- 
pered or  hardened  with  a  proportion  of  carbon  lower  than  about 
.f  of  1  per  cent.     High  carbon  steel  can  be  welded  only  imperfectly, 
and,  if  very  high  in  carbon,  not  at  all.     It  can  be  forged  with  care, 
and  cast  into  forms  as  desired.     It  can  be  tempered  or  hardened 
by  heating  to  a  full  yellow  and  quenching  in  cold  water  or  by  other 
means,  and  then  drawing  the  temper  to  the  point  desired. 

Mild  steel  should  not  be  worked  under  the  hammer  or  flanging 
press  at  a  low,  or  "  blue,"  heat,  as  such  working  is  found  in  many 
cases  to  leave  the  metal  brittle  and  unreliable.  Steel,  in  order  to 
weld  satisfactorily,  should  have  a  low  proportion  of  sulphur,  and 
special  care  is  required  in  the  operation,  because  the  range  of  tem- 
perature through  which  the  metal  is  plastic  and  fit  for  welding  is 
less  than  with  wrought  iron. 

33.  Tempering. — In  the  operation  of  tempering,  the  steel  after 
quenching  is  very  hard  and  brittle.     In  order  to  give  the  metal  the 
properties  desired,  the  temper  is  drawn  down  by  reheating  it  to  a 
certain  temperature,  and  then  quenching  again;  or  better  still,  by 
allowing    it   to    cool    gradually,    provided    the    temper    does    not 
rise  above  the  limiting  value  suitable  for  the  purpose  desired.     If 
the  reheating  is  done  in  a  bath  of  oil,  the  conditions  may  be  kept 
under  good  control  and  the  final  cooling  may  be  slow.     If  the  re- 
heating is  in  or  over  a  fire,  the  control  is  lacking  and  the  piece 
must  be  quenched  as  soon  as  the  proper  temperature  is  reached. 


244  MATERIALS  OF  ENGINEERING 

This  is  usually  determined  by  the  color  of  the  oxide  or  scale  which 

forms  on  the  brightened  surface  of  the  metal.     The  following  table 

shows  the  temperatures,,  corresponding  colors,   and  the  uses   for 

which  the  various  tempers  are  suited : 

430°  Faint  yellow  1 

450°  Straw  yellow  L  Hardest  and  keenest  cutting  tools. 

470°  Full  yellow    J 

490°  Brown  yellow  I  Cutting    tools    requiring    less    hardness    and 

or  orange    J      more  toughness. 

510°  Purplish  "1  Tools  for  working  softer  materials,  or  those 
530°  Purple  J  required  to  stand  rough  usage. 

550°  Light  blue  1  Spring  temper.  Used  for  tools  requiring 
560°  Full  blue  L  great  elasticity  or  toughness,  or  for  work- 
600°  Dark  blue  J  ing  very  soft  materials. 

34.  Special  Steels. — In  the  common  grades  of  steel,  the  valuable 
properties  are  due  to  the  presence  of  carbon  modified  in  some  degree 
by  other  ingredients,  as  already  described.     There  are  other  sub- 
stances which,  uniting  with  iron  in  small  proportions,  are  able  to 
give  to   the  combination  increased  strength,  hardness,   or   other 
valuable  properties.     We  have  thus  various  special  steels  in  which 
the  properties  may  be  due  to  the  presence  of  both  carbon  and  other 
ingredients,  or  due  chiefly  to  special  ingredients  other  than  carbon. 
Of  these  special  steels  we  may  note  the  following : 

35.  Nickel  Steel,  containing  somewhere  about  3  per  cent  of  nickel 
and  varying  amounts  of  carbon,  is  found  to  have  increased  strength 
and  toughness  as  compared  with  ordinary  steel.     Nickel  steel  is 
most  extensively  used  for  armor  plate,  though  to  some  extent  it  has 
been  employed  in  government  work  for  screw-shafts  and  for  boiler 
plates.     For  the  former  purposes  it  has  given  excellent  satisfaction, 
but  for  the  latter  use  difficulty  has  been  met  with  in  obtaining 
plates  free  from  surface  defects. 

36.  Chrome  Steel,  containing  from  0.5  to  1.5  or  2  per  cent  of 
chromium,  may  be  made  excessively  hard,  but  it  is  not  always 
reliable,  and  is  not  regarded  with  general  favor. 

37.  Tungsten  Steel  or  Mushet  Steel  is  a  steel  containing  tungsten 
and  carbon,  the  former  in  proportions  as  high  as  8  or  10  per  cent. 
This  steel  must  be  forged  with  care  and  is  excessively  hard.     The 
hardness  is  not  increased  by  tempering,  but  is  naturally  acquired 
as  the  metal  cools.     Hence  it  is  said  to  be  self-hardening.     Some 


MATERIALS  245 

specimens  contain  also  small  amounts  of  manganese  and  silver.  Its 
chief  use  is  for  lathe,  planer,  and  other  cutting  and  shearing  tools 
where  excesssive  hardness  is  required. 

38.  Uses  of  Steel  in  Marine  Construction. — In  modern  practice, 
mild  or  structural  steel  is  used  entirely  in  the  construction  of  ships. 
The  same  general  class  of  material  is  used  for  all  parts  of  boilers, 
though  the  tubes  are  still  sometimes  made  of  wrought  iron. 

Cast  steel,  as  well  as  cast  iron,  is  used  for  various  parts  of  en- 
gines, such  as  pistons,  crosshead  blocks,  columns,  bed-plates,  bear- 
ing pedestals  and  caps,  propeller  blades,  and  for  many  small  pieces 
and  fittings. 

Forged  steel  is  used  for  columns,  piston-rods,  connecting-rods, 
crank  and  line  shafting,  and  for  many  other  smaller  and  minor 
parts. 

39.  LEAD. — Lead  is  a  very  soft,  dense  metal,  grayish  in  color  after 
exposure  to  the  air,  but  of  a  bright  silvery  luster  when  freshly  cut. 
Commercial  lead  often  contains  small  amounts  of  iron,  copper, 
silver,  and  antimony,  making  it  harder  than  the  pure  metal.     It  is 
very  malleable  and  plastic.     In  engineering,  lead  is  chiefly  of  value 
as  an  ingredient  of  bearing  metals  and  other  special  alloys.     Lead 
piping  is  also  used  to  some  extent  as  suction  and  delivery  pipes  for 
water  where  the  pressure  is  only  moderate,  and  where  the  readiness 
with  which  it  may  be  bent  and  fitted  adapts  it  for  use  in  contracted 
places. 

40.  TIN. — Tin  is  a  soft,  white,  lustrous  metal  with  great  mallea- 
bility.    Commercial  tin  usually  contains  small  portions  of  many 
other  substances,  such  as  lead,  iron,  copper,  arsenic,  antimony,  and 
bismuth.     It  is  largely  used  as  an  alloy  in  the  various  bronzes  and 
other  special  metals.     Tin  resists  corrosion  well  and  in  consequence 
is  often  used  as  a  coating  for  condenser  tubes.     It  is  also  used  for 
coating  iron  plates,  the  product  being  the  so-called  "  tin  plate  "  of 
commerce.     It  melts  at  about  450°,  which  corresponds  to  a  steam 
pressure  of  about  400  Ibs.  per  square  inch.     Due  to  this  low  melting 
point,  tin  is  often  used  in  the  composition  for  safety  plugs  in  boilers. 

41.  ZINC. — Zinc,  or  "spelter,"  as  it  is  often  commercially  called, 
is  a  brittle  and  moderately  hard  metal  with  a  very  crystalline  frac- 
ture.    The  impurities  most  commonly  found  in  zinc  are  iron,  lead, 
and  arsenic.     It  is  used  chiefly  as  an  alloy  in  the  various  brasses, 
bronzes,  etc.,  and  as  a  coating  for  iron  and  steel  plates,  rods,  etc. 


246  MATERIALS  OF  ENGINEERING 

The  process  of  applying  zinc  for  such  a  coating  is  called  "  galvan- 
izing/' and  the  product  "  galvanized  "  iron  or  steel.  Electricity  is 
not  used  in  the  process,  the  articles,  after  being  well  cleaned,  being 
simply  dipped  in  a  tank  of  melted  zinc  and  then  withdrawn.  Slabs 
of  zinc  are  also  used  in  marine  boilers  to  prevent  corrosion. 

42.  ALLOYS. — A  mixture  of  two  or  more  metals  is  called  an  alloy. 
The  properties  of  an  alloy  are  often  surprisingly  different  from 
those  of  its  ingredients.     The  melting  point  is  sometimes  lower 
than  that  of  any  of  the  ingredients,  while  the  strength,  elastic 
limit,  and  hardness  are  often  higher  than  for  any  one  of  them. 

Mixtures  of  copper  and  zinc  are  called  brass.  Mixtures  of  cop- 
per and  tin,  or  of  copper,  tin,  and  zinc,  with  sometimes  other  sub- 
stances in  small  proportion,  form  gun  metals,  compositions,  and 
bronzes.  These  terms  are,  however,  rather  loosely  employed.  Vari- 
ous mixtures  of  two  or  more  of  the  metals — copper,  tin,  zinc,  lead, 
antimony — form  the  various  bearing  metals. 

Brass  and  composition  are  used  for  piping  and  pipe-fitting; 
globe,  gate,  check,  and  safety  valves;  condenser  tubes  and  shells; 
sleeves  for  tail  shafts,  and  for  a  great  number  of  small  fittings  and 
attachments  for  which  the  metal  may  be  suited.  The  bronzes  are 
employed  for  many  of  the  uses  of  brass  where  more  hardness, 
strength,  or  rigidity  are  required.  They  are  used  with  special 
success  as  a  material  for  propeller  blades. 

The  white  metals,  supported  or  backed  by  some  other  metal,  such 
as  brass,  cast  iron,  or  cast  steel,  to  give  the  necessary  strength,  are 
now  very  largely  used  for  bearing  surfaces. 

43.  TIMBER. — The  use  of  timber  in  modern  engineering  con- 
struction is  becoming  limited.     The  advance  made  in  the  produc- 
tion of  steel,  whereby  its  homogeneity  is  assured  and  its  superior 
strength  unquestioned,  has  caused  it  to  be  substituted  for  wood 
whenever  it  is  possible  and  profitable  to  do  so, 

Generally  speaking,  the  heaviest  and  darkest  colored  timber  is 
the  strongest,  and,  in  all  cases,  the  strength  of  timber  is  greatest  in 
the  direction  of  its  grain. 

The  locality  from  which  timber  comes,  the  season  of  its  cutting, 
and  the  duration  of  its  seasoning,  are  factors  in  its  strength,  and 
the  uncertainty  arising  therefrom  makes  it  advisable  to  use  a  factor 
of  safety  of  not  less  than  10  in  calculations  relating  to  the  dimen- 
sions of  timber  to  withstand  given  loads. 


MATERIALS  247 

44.  CONCRETE. — Concrete  is  a  mechanical   mixture  of  cement, 
sand,  and  broken  stone  or  gravel,  in  such  proportions  as  experiment 
has  shown  to  be  proper  for  the  conditions  governing  its  use.     It  is 
largely  used  for  laying  foundations  for  buildings  and  bridges  in  wet 
ground,  and  for  breakwaters  and  sea-walls.     After  laying,  it  soon 
hardens  to  a  strong  mass  which  is  little  permeable  to  water. 

Concrete  cannot  be  purchased  as  a  manufactured  product,  but 
must  be  made  as  needed.  Whatever  the  proportions  used,  there  is 
the  utmost  necessity  for  thorough  mixing,  such  water  being  added 
as  may  be  necessary  to  secure  coherency  in  the  mixture.  The  use 
of  machines  designed  for  the  purpose  secures  a  more  perfect  mix- 
ture than  that  attained  by  the  hand  process. 

There  is  not  perfect  agreement  as  to  the  most  advantageous  pro- 
portions for  the  composition  of  concrete,  but  good  results  may  be 
expected  from  the  proportions  of  1  of  cement  and  4  of  gravel  and 
sand,  when  used  for  columns;  if  used  for  beams,  the  proportions 
should  not  differ  much  from  1  of  cement,  2.5  of  sand,  and  4.5  of 
broken  stone. 

45.  STEEL-CONCRETE. — The  uncertainty  concerning  the  protec- 
tion from  corrosion  of  the  "  skeleton  "  steel  frame,  so  largely  used  in 
modern  building  construction,  is  mainly  responsible  for  the  very 
recent  and  extensive  use  of  "  steel-concrete  " — also  known  as  "  ferro- 
concrete "  and  as  "  reinforced-concrete  " — for  beams  and  columns 
in  the  erection  of  buildings. 

There  can  be  no  question  as  to  the  appropriate  use  of  steel  for 
constructive  purposes  in  such  open  structures  as  bridges  and  steam- 
ships, but  in  cases  where  a  steel  skeleton  is  vested  with  the  strength 
of  a  structure,  and  is  subsequently  incased  in  terra-cotta,  stone,  or 
brick,  precluding  visual  inspection,  there  is  serious  question  as  to 
the  propriety  of  its  use. 

The  use  of  concrete  for  constructive  purposes  was  common  among 
the  ancients,  and  the  fact  that  in  some  ruins,  as  they  stand  to-day, 
the  concrete  parts  remain — the  stone  having  long  since  disappeared 
— is  conclusive  evidence  of  its  durability.  There  is  abundant  evi- 
dence also  that  iron  embedded  in  concrete  is  protected  from  the 
corrosive  influence  of  moisture  and  from  the  ravages  of  fire. 

It  has  been  demonstrated  experimentally  that  concrete  is  strong 
in  compression  but  quite  weak  in  tension.  In  the  case  of  a  con- 
crete beam  supported  at  the  ends  and  loaded  at  the  middle,  it  has 
been  shown  that  the  upper  side,  which  is  in  compression,  is  capable 


248  MATERIALS  OF  ENGINEERING 

of  supporting  ten  times  the  load  which  would  cause  failure  at  the 
lower  side,  which  is  in  tension. 

Having  in  steel  a  material  of  very  high  tensile  strength,  and  in 
concrete  a  material  possessing  high  compressive  strength  as  well  as 
the  properties  of  durability  and  impermeability  to  moisture,  the 
problem  arose  of  effecting  their  combination  so  as  to  produce  a 
composite  material  having  the  desirable  qualities  of  both.  The  solu- 
tion of  this  problem  was  the  production  of  steel-concrete. 

In  the  case  of  beams,  steel  bars  are  embedded  in  the  area  of  the 
concrete  below  the  neutral  axis,  reinforcing  the  concrete  subjected  to 
tensile  stress,  and  thus  enabling  the  full  strength  of  the  compression 
area  above  the  neutral  axis  to  be  utilized  in  supporting  heavier  loads 
than  would  otherwise  be  possible. 

In  the  case  of  columns,  the  well-known  fact  that  the  most 
economical  metal  section  is  that  of  the  hollow  cylinder  suggested  at 
once  the  conception  of  a  column  having  three  concentric  parts :  (a) 
a  central  core  of  concrete;  (b)  an  outer  zone  of  concrete;  (c)  an 
intermediate  zone  of  steel.  The  steel  is  thus  protected  from  fire 
and  moisture  and  is  best  disposed  for  the  utilization  of  its  maximum 
strength. 

Between  the  tubular  reinforced  concrete  column  just  described 
and  the  plain  non-reinforced  concrete  column  there  are  a  variety 
of  possible  combinations.  In  the  form  generally  used,  applicable 
alike  to  columns  and  piles,  the  reinforcement  consists  of  vertical 
rods  of  steel  tied  together  by  a  system  of  horizontal  wires.  These 
horizontal  ties  not  only  keep  the  vertical  rods  in  position,  but  ma- 
terially assist  in  preventing  flexure  in  them;  they  also  prevent  lat- 
eral bulging  of  the  concrete. 

The  reinforcement  for  both  beams  and  columns  is  first  placed  in 
position,  after  which  they  are  enveloped  by  a  wooden  form,  or 
mould,  into  which  the  concrete  is  dumped,  and  rammed  when 
necessary.  After  a  period  of  thirty  hours  the  form  may  be  removed 
and  the  steel-concrete  product  allowed  to  season  for  several  weeks 
before  being  subjected  to  its  load. 

46.  Adhesion  of  Concrete  to  Steel. — It  has  been  shown  by  experi- 
ment that  the  concrete  on  the  compression  side  of  steel-concrete 
beams  may  be  subjected  without  rupture  to  a  stress  twenty  times 
that  which  would  cause  failure  in  a  tension  test  in  the  concrete 
alone.  It  has  also  been  shown  that  reinforced  concrete  acquires 


MATERIALS  249 

a  power  to  resist  crushing  which  is  greater  than  the  sum  of  the 
resistances  of  the  two  materials  taken  separately.  Such  remarkable 
results  could  not  be  obtained  were  it  not  for  the  adhesion  between 
the  concrete  and  steel,  such  adhesion  offering  resistance  to  sliding 
between  the  two  surfaces  and  facilitating  the  transference  of  the 
forces  from  one  surface  to  the  other. 

The  bond  between  the  steel  and  concrete  occasioned  by  adhesion 
alone  is  liable  to  be  destroyed  by  internal  stresses  due  to  shocks  and 
vibrations,  and  from  unequal  expansions  resulting  from  thermal 
changes,  and  it  is  with  the  idea  of  strengthening  the  adhesion  bond 
by  mechanical  means  that  the  reinforcing  bars  are  sometimes 
twisted  or  corrugated. 

47.  Distinction  between  "  Steel  and  Concrete  Construction  "  and 
"  Steel-Concrete  Construction." — In  steel  and  concrete  construction, 
beams  and  columns  of  large  section  form  the  basis  of  strength 
of  the  structure,  the  concrete  contributing  no  assistance,  its  weight 
in  fact  forming  a  part  of  the  load  borne  by  the  steel  members.  In 
steel-concrete  construction,  the  reinforcing  steel  takes  care  only  of 
the  tensile  stress  in  beams — leaving  the  entire  compressive  stress 
to  be  borne  by  the  concrete — while  in  columns  it  increases  the  com- 
pressive strength  and  tends  to  prevent  bulging  of  the  concrete.  To 
secure  a  uniform  distribution  of  the  stress  in  the  reinforced  section 
of  concrete,  the  reinforcement  should  consist  of  steel  bars  of  small 
section  distributed  so  that  each  shall  bear  its  allotted  part  of  the 
stress.  The  proportion  of  steel  to  concrete  depends  directly  upon 
the  ratio  of  the  coefficients  of  elasticity  of  the  two  materials.  The 
value  of  E  for  steel  is  30,000,000  Ibs.  per  sq.  inch,  implying  that  a 
force  of  one  pound  would  extend  or  compress  a  bar  1  sq.  inch  in 

area  by  QQ  QQQ  QQQ  =  0.000000033  of  its  original  length.  The 
value  for  concrete  may  be  taken  as  3,000,000,  so  that  for  an  equal 
extension  of  the  two  materials  the  steel  will  bear  ten  times  the 
stress  that  can  be  borne  by  the  concrete. 

Suppose  a  bar  of  steel  1  sq.  inch  in  area  to  be  surrounded  by  a 
ring  of  concrete  1  sq.  inch  in  area;  and  suppose  further  that  the 
steel  be  subjected  to  a  direct  pull  that  would  occasion  in  the  con- 
crete the  safe  allowable  tension  stress  of  50  Ibs.  per  sq.  inch.  The 

elongation  in  the  concrete  would  then  be  50  X  3  QQQ  QQQ 
=  0.0000167  inch.  Considering  the  bond  between  the  steel  and 


250 


MATERIALS  OF  ENGINEERING 


concrete  to  be  perfect,  the  steel  would  suffer  an  equal  elongation, 
occasioning  a  stress  in  the  steel  of  0.0000167  X  30,000,000  =  500 
Ibs.  per  sq.  inch,  a  result  only  one-thirtieth  of  the  safe  allowable 
tension  stress  for  steel;  consequently,  the  sectional  area  of  the 
steel  may  be  reduced  to  -^  sq.  inch,  thus  raising  its  stress  to  15,000 
Ibs.  per  sq.  inch  without  causing  failure  in  the  concrete. 

48.  Design  of  Steel-Concrete  Beam, — When  a  beam  is  deflected 
the  section  above  the  neutral  axis  is  in  compression  and  that  below 
the  neutral  axis  is  in  tension  (see  Art.  51,  p.  312).  The  tension 
is  not  the  result  of  a  direct  pull,  and  in  consequence  the  unit  stress 
is  not  the  same  throughout  the  section  below  the  neutral  axis  but 
varies  directly  as  the  distance  from  the  neutral  axis. 


\ 


In  the  case  of  a  steel-concrete  beam  the  tensile  stress  will  be  taken 
both  by  the  steel  and  concrete,  but  as  the  bond  is  assumed  to  be 
perfect  the  concrete  will  be  permitted  to  extend  only  to  the  same 
amount  as  the  steel.  The  amount  of  the  total  stress  borne  by  the 
concrete  will  depend  upon  the  proportion  of  steel  to  concrete. 

In  a  steel  beam  the  resistance  to  tension  and  to  compression  are 
equal  and  the  neutral  axis  passes  through  the  center  of  gravity  of 
the  section,  but  in  the  case  of  a  concrete  beam  where  the  compres- 
sion strength  of  the  material  is  much  greater  than  its  tension 
strength,  the  neutral  axis  moves  away  from  the  axis  through  the 
center  of  gravity  as  the  stress  increases.  , 

The  formulae  for  the  deflection  of  beams  are  based  on  the  sup- 
position that  the  neutral  axis  passes  through  the  center  of  gravity 
(see  Part  IV),  so  that  some  modification  will  be  necessary  to  make 
them  applicable  to  steel-concrete  beams. 

Knowing  that  the  ratio  of  the  compressive  and  tensile  strengths 
of  concrete  is  10  :  1,  we  may  inquire  as  to  the  probable  location  of 
the  neutral  axis  in  a  concrete  beam. 


MATERIALS 


251 


Let  oo,  Fig.  11,  denote  the  position  of  the  neutral  axis  of  the 
section  of  a  rectangular  concrete  beam;  c  and  cv  the  distances 
from  the  neutral  axis  of  the  outermost  fibers  of  the  compression 
and  tension  areas  of  the  section  respectively;  li  and  ht  the  dis- 
tances from  the  neutral  axis  of  the  centers  of  pressure  of  the  com- 
pression and  tension  areas  respectively;  and  &  the  breadth  and  d 
the  depth  of  the  section. 

The  total  stress  in  a  section  is  the  product  of  the  mean  stress  act- 
ing at  the  center  of  pressure  and  the  whole  area,  and  the  moment 
of  this  stress  with  respect  to  the  neutral  axis  may  be  taken  as  the 
measure  of  the  resistance  of  the  section  to  bending. 

The  compressive  working  stress  of  the  concrete  being  10  times 
the  tensile  working  stress,  with  equal  deformations,  the  mean  com- 


Y 

X 

t 

i 

c/V 

**  h    *• 

Fi9.  IZ. 

pressive  stress  will  be  denoted  by  10$  when  S  denotes  the  mean 
tensile  stress.  The  resistance  to  bending  offered  by  the  compres- 
sion area  will  then  be  lOSbch,  and  that  offered  by  the  tension  area 
will  be  SbcJi-L.  Since  these  resistances  must  be  equal  for  equilib- 
rium, we  shall  have : 

WSbch  =  S1)cji19  whence  Wch  =  cj^.  (1) 

The  center  of  pressure  of  a  rectangle  on  which  the  pressure  in- 
creases uniformly  from  0  at  one  edge  to  p  at  the  opposite  edge  may 
be  found  as  follows : 

In  the  rectangle  of  breadth  &  and  altitude  h,  Fig.  12,  we  have : 
Elemental  area  =  bdy, 

Elemental  pressure  =  bdy  X  ^  , 
Whole  pressure  =  -P-  I  ydy, 
Moment  of  whole  pressure  ==-?•/  y*dy . 


252  MATERIALS  OF  ENGINEERING 

The  distance  y  of  the  center  of  pressure  from  the  edge  where  the 
pressure  is  0  is  then, 

»n  y*~\h 

y*dy   a  1    2h 

3  ' 


tf5    I 

fi 


Then,  in  Fig.  11,  we  shall  have  li  =  ~,  and  ht  =  2|^.     Substi- 

tuting these  values  of  c  and  c±  in  (1),  we  have  : 

20?2      2<2 

—  Q-  =  -«*-,  whence  ^  =  c\/10. 

But  c  +  c±  =  c  +  cVIO  —  d,  whence  c  =  0.24d,  and  c±  =  0.76d, 
which  fixes  the  position  of  the  neutral  axis  under  our  suppositions. 

Conversely,  if  sufficient  steel  be  added  to  the  tension  area  of  the 
section  to  make  the  resistance  of  steel  +  the  resistance  of  concrete 
ten  times  greater  than  the  resistance  of  the  concrete  in  the  compres- 
sion area  of  the  section,  we  shall  derive,  c  =  0.76^  and  c±  =  0.24d 

From  this  we  infer  that  the  position  of  the  neutral  axis  may  be 
made  to  depend  upon  the  proportions  of  concrete  and  steel,  and 
upon  the  disposition  of  the  steel  relative  to  the  neutral  axis.  We 
may,  therefore,  determine  the  position  of  the  neutral  axis  by  calcu- 
lation when  these  details  have  been  settled,  or  we  may  determine 
the  proportions  of  steel  and  concrete  for  any  chosen  position  of  the 
neutral  axis. 

EXAMPLE.  —  A  steel-concrete  beam  of  rectangular  section  is  to 
have  a  clear  span  of  10  feet  and  support  a  uniformly  distributed 
load  of  6000  Ibs.  Assuming  the  depth  of  the  beam  to  be  14  inches, 
and  that  the  neutral  axis  is  9  inches  from  the  lower  edge  of  the  beam, 
it  is  required  to  determine  the  breadth  of  the  beam  and  the  sectional 
area  of  the  reinforcement. 

From  Art.  30,  p.  291,  we  have,  for  a  section  in  which  the  neutral 

WT 
axis  passes  through  the  center  of  gravity,  Bending  Moment  =  —^  ; 

*  c  7" 

and  from  Art.  53,  p.  313,  we  have,  Eesisting  Moment  =  —  .     For  a 
rectangular  section,  I  =  yo  ,  and  we  shall  have,  for  equilibrium, 


since  d  =  2c. 
Whence, 


WL      Sbd3 

"' 


MATERIALS  253 

in  which  8  is  the  extreme  fiber  stress.     The  mean  unit  stress  be- 
tween the  neutral  axis  and  the  outermost  fiber  is   -  —I —  =  ^  , 

and  the  total  stress,  F,  is  therefore,   be   X    o  >  from  which  we 
derive : 

F-3WL  U\ 

•w 

Considering  the  reinforcement  to  be  placed  in  the  axis  passing 
through  the  center  of  pressure,  and  the  whole  of  the  tensile  stress 
to  be  concentrated  in  the  reinforcement,  we  have,  by  substitution 
in  (4)  : 


32  X  9 


l 


Since  the  safe  allowable  stress  for  steel  is  15,000  Ibs.  per  sq.  inch, 
the  requisite  sectional  area  of  the  steel  reinforcement  is, 


Taking  the  safe  allowable  compressive  fiber  stress  in  the  concrete 
as  500  Ibs.  per  sq.  inch,  we  have,  from  (2)  : 

_  3WL  _  3  X  6000  X  120  _  h 

~'~~  16X500  X25  ~ 


as  the  width  of  the  concrete. 

The  dimensions  of  the  beam  will  then  be  14  inches  in  depth  by 
10.8  inches  width,  making  10.8  X  14  —  151.2  sq.  inches  area  of 
section. 

A  section  of  the  beam  is  shown  in  Fig.  13,  the  reinforcement  con- 
sisting of  two  bars,  each  of  0.25  sq.  inch  section,  placed  in  the  axis 


254  MATERIALS  OF  ENGINEERING 

through  the  center  of  pressure  in  the  tension  area,  distant  3  inches 
from  the  bottom  of  the  beam. 

This  treatment  of  steel-concrete  is  neither  satisfactory  nor  con- 
clusive. Indeed,  the  use  of  this  composite  material  has  not  yet 
emerged  from  the  experimental  state,  and  the  opinion  is  entertained 
by  some  engineers  that  thirty  years  hence  the  now  popular  theories 
concerning  both  steel  and  concrete  will  have  undergone  changes. 


CHAPTER  II. 
TESTING  MATERIALS. 

49.  Stress. — The  application  of  external  forces  to  a  piece  of  ma- 
terial tends  to  change  its  shape,  and  this  tendency  induces  internal 
forces,   known  as  stresses,  which  offer  resistance  to  the  change. 
These  stresses  may  be  of  three  kinds : 

(1)  If  the  external  force  be  applied  at  right  angles  to  the  sec- 
tion, and  acts  away  from  it,  the  stress  is  one  of  tension,  or  a  tensile 
stress. 

(2)  If  the  external  force  acts  toward  the  section,  the  stress  is 
one  of  compression,  or  a  compressive  stress. 

(3)  If  the  external  force  acts  parallel  to  the  section,  the  stress  is 
one  of  shear,  or  a  shearing  stress. 

It  is  a  fundamental  assumption  that  these  direct  stresses  are  uni- 
formly distributed  over  the  section,  so  that  if  F  denotes  the  external 
force,  or  load,  A  the  area  of  section,  and  8  the  unit  stress,  we  must 
have,  in  the  absence  of  rupture,  F  =  AS.  F  is  usually  expressed  in 
pounds  and  A  in  square  inches,  so  that  we  shall  have  for  the  unit 
stress : 

«       F 
-J 

in  pounds  per  square  inch. 

50.  Strain. — A  piece  of  material  which  is  stressed  by  the  appli- 
cation of  external  force  undergoes  some  change  in  its  dimensions, 
either  lengthened  or  shortened,  and  the  amount  of  this  distortion  is 
known  as  the  strain  due  to  the  external  force,  or  load. 

51.  Different  Kinds  of  Tests. — Materials  are  tested  for  tension, 
by  pulling  apart  a  test  piece  of  specified  dimensions ;  for  compres- 
sion by  crushing  a  piece  of  definite  dimensions;  in  cross-breaking, 
by  supporting  a  piece  at  two  points  and  breaking  or  bending  it  in  a 
testing  machine  by  applying  a  load  at  an  intermediate  point;  in 
torsion,  by  twisting  apart  a  piece  in  a  machine  designed  for  the 
purpose;  in  direct  shearing,  by  breaking  a  riveted  or  pin-joint  con- 
nection in  a  machine ;  for  impact,  or  shock,  by  letting  a  weight  drop 
through  a  definite  height,  and,  by  its  blow,  develop  suddenly  the 
stress  in  the  material. 


256  MATERIALS  OF  ENGINEERING 

52.  Ultimate  Strength. — The  ultimate  strength  of  a  test  piece  is 
the  load  required  to  produce  fracture,  reduced  to  a  square  inch  of 
original  section;    or,  in  other  words,  it  is  the  ultimate  or  highest 
load,  divided  by  the  original  area.     Thus,  if  the  area  of  the  cross- 
section  of  a  test  piece  is  0.42  sq.  inch,  and  the  load  producing 

fracture  is  28,400  Ibs.,  the  ultimate  strength  is  ^'||0  =  67,620 
Ibs.  per  sq.  inch. 

53.  Factor  of  Safety. — The  factor  of  safety  is  the  quotient  aris- 
ing from  dividing  the  ultimate  strength  by  the  unit  stress. 

54.  Elastic  Limit. — The  elastic  limit  is  the  load,  per  square  inch 
of  area,  which  will  just  produce  a  permanent  set  in  the  material. 
Thus,  in  a  tension  test,  if  the  cross-sectional  area  of  the  test  piece 
be  0.7  square  inch,  and  a  permanent  set  just  be  produced  by  a  load 
of  28,000  Ibs.,  the  elastic  limit  is  40,000  Ibs.  per  sq.  in. 

55.  Elongation. — The  increase  in  length  of  a  test  piece,  measured 
just  before  rupture,  divided  by  the  original  length  of  the  piece,  ex- 
presses the  elongation. 

When  a  load  is  first  applied  to  a  test  piece,  the  elongation  is 
nearly  uniformly  distributed  throughout  the  whole  length.  This 
continues  until  the  piece  begins  to  contract  in  area  near  the  point 
of  final  rupture,  and  nearly  all  the  subsequent  elongation  is  re- 
stricted to  the  immediate  vicinity  of  this  point. 

In  expressing  the  elongation  of  any  material,  the  length  of  the 
test  piece  must  be  stated.  The  usual  length  of  test  piece  is  8 
inches,  so  that  if  an  extension  in  length  of  2  inches  were  noted  just 
before  rupture,  the  elongation  would  be  expressed  as  25  per  cent  in 
8  inches. 

56.  Reduction  in  Area. — The  reduction  of  area  is  found  by  divid- 
ing the  difference  between  the  original  and  final  sectional  areas  at 
the  point  of  rupture  by  the  original  area,  expressing  the  fraction  in 
per  cent. 

57.  Test  Piece  for  Iron. — The  form  of  test  piece  for  wrought  iron 
plate  prescribed  by  the  U.  S.  Board  of  Supervising  Inspectors  of 
Steam  Vessels  is  illustrated  in  Fig.  1.     If  the  plate  is  y^mcn  thick, 
or  less,  the  width  at  the  reduced  section  must  be  1  inch.     If  the 
plate  is  over  T5¥  inch  in  thickness,  the  width  of  the  piece  must  be 


TESTING  MATERIALS 


257 


reduced  so  that  the  cross-sectional  area  at  the  reduced  section  shall 


to* 


?.  /.  Test  F<*ce  for  Iron, 

be  about  0.4  sq.  inch,  but  it  must  not  be  greater  than  0.45  sq.  inch, 
nor  less  than  0.35  sq.  inch. 

58.  Test  Pieces  for  Steel  and  Other  Materials. — Figure  2  shows 


*0 

i 


?.  Trarf  Piece  /or  Jteet. 

the  form  of  test  piece  for  tension  prescribed  by  the  Navy  Department 
for  tests  of  steel  plates  for  naval  uses. 

Figure  3  shows  the  form  prescribed  by  the  Association  of  Ameri- 

V£-  /  RADIUS  J 

fc ^ 


r 


^ 


can  Steel  Manufacturers,  and  adopted  by  the  U.  S.  Board  of  Super- 
vising Inspectors.   The  test  piece  for  plates  is  cut  from  a  "  coupon/' 


fig.  -4  •  Plate  toit/i    Coupon. 

as  it  is  called,  left  on  one  corner  of  the  plate  as  shown  at  A,  Fig.  4. 
The  U.  S.  law  requires  further  that : 
17 


258 


MATERIALS  OF  ENGINEERING 


"  Every  iron  or  steel  plate  intended  for  the  construction  of  boil- 
ers to  be  used  on  steam  vessels  shall  be  stamped  by  the  manufac- 
turer in  the  following  manner:  At  the  diagonal  corners,  at  a  dis- 


8* 

Round  T<ut  Pi**. 

tance  of  about  4  inches  from  the  edges  and  at  or  near  the  center 
of  the  plate,  with  the  name  of  the  manufacturer,  the  place  where 
manufactured,  the  number  of  pounds  tensile  stress  it  will  bear 
to  the  sectional  inch." 


ij.  6.  Bending  Test 

Figure  5  shows  the  usual  round  form  of  test  piece  for  all  materials 
except  plates. 


7.  Jh^ft 

Figure  6  illustrates  a  cold  bending  test  on  a  piece  of  steel  plate. 
A  drift  test  is  sometimes  required.   This  is  illustrated  in  Fig.  7,  and 

e. 


uu 

r  Jntf*  Test. 

consists  in  driving  tapered  drifts  of  different  sizes  into  a  punched 
or  drilled  hole  until  the  diameter  is  increased  to  at  least  twice  its 


TESTING  MATERIALS 


259 


original  size.     The  metal  must  stand  this  test  without  sign  of  frac- 


ture about  the  edges  of  the  hole. 
T  irons  are  illustrated  in  Fig.  8. 


Bending  tests  for  angle  and 


59.  Ductility. — Materials  such  as  wrought  iron,  mild  steel,  cop- 
per, and  other  metals  which  may  be  lengthened  by  the  application 
of  an  external  force,  with  a  corresponding  decrease  in  thickness  or 
in  diameter,  possess  the  property  of  ductility. 

60.  Plasticity. — If,  in  the  process  of  stretching  a  ductile  mate- 
rial, the  load  be  removed  and  none  of  the  strain  disappears,  the  ma- 


Fig.  9. 

• 

terial  is  said  to  have  passed  beyond  the  ductile  and  to  have  entered 
the  plastic  stage.  The  plasticity  of  the  material  is  determined  by 
the  final  elongation  and  contraction  in  area  of  the  test  piece.  In 
structures  subjected  to  live  loads  and  shocks  it  is  equally  important 
to  know  the  power  of  the  material  used  to  resist  deformation  with- 
out rupture  as  it  is  to  know  the  ultimate  strength,  and  for  that 
reason  specifications  for  iron  and  steel  usually  require  a  certain 
percentage  of  elongation  and  contraction  of  area  in  a  stated  length 
of  test  piece. 

61.  Stress-Strain  Diagram. — If,  in  testing  a  material,  the  gradu- 
ally applied  loads  be  plotted  as  ordinates  and  the  corresponding 


260  MATERIALS  OF  ENGINEERING 

strains  as  abscissas,  the  resulting  curve  is  known  as  a  stress-strain 
diagram. 

For  tension  tests  of  wrought  iron  and  mild  steel  such  a  diagram 
will  take  the  form  shown  in  Fig.  9. 

The  load  being  gradually  increased,  it  will  be  found  that,  within 
a  certain  limit,  the  strains,  or  extensions,  will  be  directly  propor- 
tional to  the  augmentations  in  the  load,  and  that,  if  the  stress 
be  relieved,  the  test  piece  will  return  to  its  original  length.  Dur- 
ing this  period  the  material  is  said  to  be  perfectly  elastic,  and  will 
be  so  represented  in  the  diagram  by  the  straight  line  OA .  By  con- 
tinuing the  gradual  increase  in  the  load  a  point  will  be  reached 
where  the  proportionality  between  the  strain  and  the  augmenta- 
tion in  the  load  ceases,  the  strain  increasing  much  more  rapidly 
than  the  load ;  and  if  the  stress  be  relieved  the  piece  will  not  return 
to  its  original  length,  but  will  acquire  a  permanent  set.  The  load 
at  which  this  occurs  is  known  as  the  elastic  limit  of  the  material. 

A  further  increase  in  the  load  very  soon  develops  a  point  where 
the  extension  increases  very  rapidly — as  much  as  from  10  to  15 
times  its  previous  amount — known  as  the  yield  point.  This  rapid 
increase  in  the  extension  usually  occasions  an  apparent  reduction 
in  the  stress,  as  shown  by  the  fall  in  curvature  beyond  the  point  B. 
For  commercial  purposes  the  yield  point  and  the  elastic  limit  are 
taken  as  the  same  point,  and  the  ordinate  at  B  would  represent,  to 
the  scale  of  the  diagram,  this  limit  in  pounds  per  square  inch  of 
section  of  the  material. 

Passing  the  yield  point  the  strains  increase  much  faster  than  the 
loads,  but  if  the  stress  in  ftie  material  be  relieved  a  careful  meas- 
urement will  show  the  disappearance  of  a  small  portion  of  the 
extension,  indicating  the  existence  still  of  some  elasticity  and  that 
the  specimen  is  passing  through  the  ductile  stage. 

At  about  the  time  the  maximum  stress  is  reached  at  (7,  the  mate- 
rial appears  to  have  reached  the  plastic  state,  the  extension  in- 
creasing, in  time,  without  increase  in  load.  Up  to  this  point  the 
strain  has  been  evenly  distributed  throughout  the  length  of  the 
specimen,  but  here  occurs  an  extension  and  reduction  in  section 
purely  local,  immediately  followed  by  rupture  at  D. 

Within  the  elastic  limit  the  extension  of  the  specimen  probably 

would  not  exceed  of  its  length,  so  it  is  quite  impossible  to 


TESTING  MATERIALS 


261 


make  direct  measurements  of  the  extensions  corresponding  to  the 
augmentations  in  load.  Some  form  of  extensometer  is  used  to 
make  these  measurements. 

62.  Tension  Test  of  Mild  Steel. — For  example,  Fig.  10  is  a  stress- 
strain  diagram  for  a  tensile  test  of  mild  steel. 


So  ooo 


2.6000 


* 


I 

$ 


2.2.000  \-r~ 

2oooo 

/Qooo 


/jooo 


IOOOO 


6*00 


2.000 


s* 

*^- 

\ 

^ 

X 

^ 

/ 

/ 

* 

£ 

57/> 

fee 

•/« 

Z7 

"9 

ra 

m 

r^7 

7i 

»/75 

/o/- 

r    7 

?5-7 

c  01 

>ee 

/"  2' 

£xfenstons.     Sca/e^  fc///$/ze. 

ng.  /o 


The  original  dimensions  of  the  test  piece  were : 

Length,  8  inches;  diameter,  0.75  inch;  area  of  section,  0.4418 
square  inch. 

The  final  dimensions  were : 

Length,  10.2  inches;  diameter,  0.4843  inch;  area  of  section, 
0.1842  square  inch. 

The  load  was  gradually  applied  with  the  uniform  augmentation 
of  2000  pounds,  and  the  data  tabulated  as  follows : 


262 


MATERIALS  OF  ENGINEERING 


Loads. 

Extensions. 

Remarks. 

2,000 

0.0012 

4,000 

0  .  0024 

6,000 

0.0034 

8,000 

0.0047 

10,000 

0.0060 

12,000 

0.0072 

14,000 

0.0083 

16,000 

0.0095 

18,000 

0.0106 

20,000 

0.0118 

22,000 

0.0220 

Yield  Point 

26,000 

0  .  5300 

29,680 

1  .  9200 

Maximum  Stress 

25,820 

2.2000 

Breaking  Stress 

An  inspection  of  the  diagram  shows  the  elastic  limit  to  have 
been  reached  at  about  21,000  pounds. 

t>  I  fifin 

Elastic  limit  of  specimen  =  Q  4418  =  ^>^®  ^s.  Per  scl-  incw- 

maximum  load      29,700 
Ultimate  stress  =  -  — —  =  ^-^-a  rs  67/220  Ibs.  per  sq. 


inch. 


original  area    "    0.4418 


„  ,      .       -     Q  •     u  (10.2  -  8)100        0_,  K 

Extension  in  8. inches  =  — g— j      -  =  27.5  per  cent. 

(0.4418  -  0.1842)100 
Contraction  of  area  =  04418  ~  58.31  per  cent. 

To  find  the  modulus  of  elasticity  we  proceed  as  follows : 
The  sum  of  the  extensions  up  to  and  including  the  18;000  pound 
load — a  point  well  within  the  limit  of  elasticity — is  0.0533  inch, 
and  the  sum  of  the  loads  is  90,000  pounds.     The  mean  extension 

0  0533 
for  2000  pounds  is,  therefore,    ' ^      —  0.0012  inch. 

T  ^ 

By  Art.  50,  page  311,  we  have,  E  =  -  — ,  in  which  L  is  the  orig- 


TESTING  MATERIALS 


263 


264  MATERIALS  OF  ENGINEERING 

inal  length  and  8  the  load  producing  the  extension   y.     Here 
L  =  8  inches,  S  =  2000  pounds,  and  y  =  0.0012. 

Hence>  E  =  =  30>180>000  lbs-  per  s(iuare  inch- 


63.  Compression  Tests.  —  In  testing  materials  for  compression  the 
specimens  are  not  longer  than  from  1.5  to  3  times  the  diameter.  If 
the  specimens  are  long  the  failure  under  compression  will  be  by 
buckling  or  bending,  and  for  intermediate  lengths  partly  by  crush- 
ing and  partly  by  bending. 


PART  IV 

THE  ELEMENTS  OF  THE  MECHANICS 
OF  MATERIALS 


CHAPTER  I. 
MOMENTS.— CENTER  OF  GRAVITY. 

1.  Moments. — The  moment  of  a  force  acting  on  a  body  may 
be  defined  as  the  power  of  the  force  to  turn  the  body  about  a  point, 
or  about  a  fixed  axis,  and  its  measure  is  the  product  of  the  force  by 
the  perpendicular  distance  from  the  point,  or  from  the  axis,  to 
the  line  of  action  of  the  force.  The  point,  or  axis,  about  which 
the  moments  are  taken  is  called  the  center  of  moments. 

If  there  be  a  number  of  forces  acting  on  the  body,  those  tending 
to  turn  it  in  one  direction  may  be  regarded  as  positive,  and  those 
tending  to  turn  it  in  the  opposite  direction  as  negative.  By  com- 
mon consent  forces  with  a  turning  tendency  in  a  clockwise  direction 
are  termed  positive  and  those  with  a  contra-clockwise  tendency 


negative.  It  is  immaterial  which  kind  of  force  is  termed  positive 
and  which  negative,  but  having  chosen  one  kind  as  positive  in  any 
investigation  the  choice  must  be  adhered  to,  and  the  opposite  kind 
must  be  regarded  as  negative. 

2.  In  Fig.  1,  let  the  forces,  P,  Q,  and  R  act  on  a  body  in  the 
directions  indicated.  If  the  body  remains  in  equilibrium  the  under- 
lying principle  of  moments  asserts  that  the  algebraic  sum  of  the 
moments  of  the  forces  about  any  point  as  a  center,  or  about  any 
line  as  an  axis — the  point  and  the  line  being  in  the  same  plane — 
must  be  zero.  In  other  words,  the  clockwise  moments  must  be 
equal  to  the  contra-clockwise  moments. 

Let  moments,  be  taken  about  the  point  0.  The  force  Q  tends  to 
turn  the  body  in  a  clockwise  direction  about  0  and  will  be  regarded 


268        THE  ELEMENTS  OF  THE  MECHANICS  OF  MATERIALS 

as  positive,  and  the  tendency  of  the  forces  P  and  K  is  to  turn  it 
in  a  contra-clockwise  direction  about  0  and  will  he  regarded  as 
negative.  The  equation  of  moments  will  then  be,  Qq  —  Pp  —  Rr 
=  0.  This  follows  directly  from  the  meaning  of  the  word  equilib- 
rium, which  implies  that  the  body  is  at  rest,  and  this  condition  can 
only  result  when  there  is  no  tendency  to  turn  the  body  about  the 
point  0 ;  that  is,  when  the  algebraic  sum  of  the  moments  about  0 
is  zero.  Should  the  line  of  action  of  a  force  pass  through  the  center 
of  moments,  the  moment  of  that  force  would  vanish. 

The  principle  of  moments  is  very  broad  in  its  applications,  and 
if  we  denote  a  force  by  /,  a  mass  by  m,  an  area  by  a,  a  volume  by  v, 
and  the  arm  of  each  by  r,  then  fr,  mr,  ar,  vr  will  be  the  moment 
of  the  force,  mass,  area,  and  volume,  respectively. 

In  expressing  the  value  of  moments,  the  units  of  force,  mass, 
area,  and  volume  are  placed  first,  and  the  length  units  afterward. 
For  example,  a  moment  may  be  expressed  as  so  many  pounds-feet, 
and  thus  avoid  confusion  with  work  units. 


3.  Center  of  Gravity.  —  The  center  of  gravity  of  a  body  or  of  a 
system  of  bodies  is  a  point  on  which  the  body  or  system  will  balance 
in  all  positions,  supposing  the  point  to  be  supported,  the  body  or 
system  to  be  acted  on  only  by  gravity,  and  the  parts  of  the  body  or 
system  to  be  rigidly  connected  to  the  point. 

It  follows  from  this  definition  that,  all  the  particles  of  a  body  or 
system  of  bodies  are  acted  on  by  a  system  of  parallel  forces  —  gravity 
acting  on  each  —  and  that  the  algebraic  sum  of  the  moments  of  these 
forces  about  a  line  must  be  zero  when  the  line  passes  through  the 
center  of  gravity  of  the  body  or  system  ;  otherwise  the  body  or  sys- 
tem would  not  balance. 

4.  Let  two  heavy  bodies  of  weights  P  and  Q  be  situated  as  shown 
in  Fig.  2.     Join  them  with  a  straight  line  and  divide  it  at  C,  so 


-Q  =  -. 

x>  r 

weight,  the  system  will,  by  the  principle  of  the  lever,  balance  when 


that  -Q  =  -.     If  the  weights  be  joined  by  a  rigid  rod  without 

x>  r 


MOMENTS. — CENTER  OF  GRAVITY.       .  269 

supported  at  C.  C  is,  therefore,  the  center  of  gravity  of  the  system. 
As  the  resultant  of  the  weights  is  P  -\-  Q,  the  pressure  on  the  sup- 
port will  be  P  +  Q.  The  center  of  gravity  of  a  uniform  straight 
rod  is,  evidently,  at  the  middle  point  of  its  length.  When  consid- 
ering a  body  at  rest,  we  may  assume  its  whole  mass  to  be  concen- 
trated at  the  center  of  gravity. 

5.  Let  the  weights  P  and  Q,  Fig.  3,  be  attached  as  shown  to  a 
balanced  rod.     Then,  as  has  just  been  shown,  the  center  of  gravity 

will  be  so  situated  that  7^  =  — ,  or  Pm  —  Qn\  that  is,  the  algebraic 

\J  nt 

sum  of  the  moments  is  zero.  In  locating  the  position  of  the  center 
cf  gravity  of  a  system  it  will  be  convenient  to  take  moments  about 
a  point  other  than  the  center  of  gravity,  or  about  a  line  other  than 
one  passing  through  the  center  of  gravity.  Let  x  denote  the  dis- 
tance of  the  center  of  gravity  of  the  system  from  0,  and  let  p  and  q 
be  the  distances,  respectively,  of  the  centers  of  gravity  of  P  and  Q 


Fig.  3. 


from  0.     We  have  from  the  figure  ra  =  x  —  p,  and  n  =  q  —  x. 
Then  P(x  —  p)  =Q(q  —  x),  from  which  we  have  x  =    p  ,Q   • 

T-  ,,  i  ,  ,    -       Pp  +  Qq  +  Rr  +  &c. 

If  the  system  be  extended  the  result,  x  =      p      ^       „       «  — 

will  be  obtained.     This  formula  is  extensively  used  in  the  solution 
of  problems. 

6.  In  considering  a  sheet  of  uniform  thickness  weighing  M 
pounds  per  unit  of  area,  the  weight  of  any  given  area  will  be  Ma 
pounds.  We  may  substitute  Ma^  for  P  and  Ma2  for  Q  in  the  for- 
mula and  obtain 


-  _ 


+  «2  av  +  a* 

in  which  A  is  the  whole  area. 
This  may  be  expressed  as  follows: 


270        THE  ELEMENTS  OF  THE  MECHANICS  OF  MATERIALS 

'Distance,  x,  from  0  of  the  center  of  gravity  is 

sum  of  the  moments  of  all  the  elemental  surfaces  about  0 

area  of  surface 
the  moment  of  the  whole  surface  about  0 


whole  area  of  surface 
Similarly, 

the  moment  of  the  whole  volume  about  0 


x  = 


whole  volume 

The  moment  of  the  whole  surface  and  of  the  whole  volume  is  ob- 
tained by  integration. 

7.  The  center  of  gravity  of  a  plane   figure   may  be   obtained 
graphically  as  follows : 


Let  dbcde,  Fig.  4,  be  any  plane  figure.  Draw  eb  and  ec.  Let 
A^  and  g19  A2  and  g2,  and  As  and  g&  denote  the  areas  and  centers 
of  gravity  of  the  triangles  eab,  ebc,  and  ecd,  respectively.  Join  g^ 
and  g2.  The  center  of  gravity  of  the  figure  abce  must  lie  on  this 
line.  From  g^  lay  off  in  any  direction  and  to  any  scale,  the  dis- 
tance g^t  equal  to  A29  and  in  the  opposite  direction,  and  to  the  same 
scale,  lay  off  from  g2,  parallel  to  g^t,  the  distance  g.2s  equal  to  A^. 
Join  st,  and  the  intersection,  g19  2,  of  st  and  g^gz  is  the  center  of 
gravity  of  the  figure  abce.  Join  glt  2  with  g3,  and  the  center  of 
gravity  of  A^  -(-  A2  +  A3  will  lie  in  this  line.  From  #1?  2  lay  off  a 
distance  gly  2r  equal  to  the  area  AM  and  from  g^  lay  off  gav  parallel 
to  <71?  zr  and  equal  io  A^-\-  A2.  Join  vr,  and  its  intersection  with 
ffi>  2^3  gives  ^u  2?  3  as  tne  center  of  gravity  of  the  whole  figure. 


MOMENTS. — CENTER  OF  GRAVITY. 


8.  Should  the  surface  contain  a  hole,  as  fmn  in  Fig.  5,  we  would 
proceed  as  follows : 

Denote  by  At  and  gly  and  A2  and  g2  the  areas  and  centers  of 
gravity  of  the  whole  figure  abcde  and  of  fmn,  respectively.  Join 
g^g2.  From  g2  lay  off  to  scale,  in  any  direction,  g2s  equal  to  Alf 
and  parallel  to  it,  and  on  the  same  side  of  g^g2,  lay  off  g^r.  Join 


sr  and  produce  it  to  meet  g^g2  produced  at  gl9  2.     Then  glf  2  is  the 

required  center  of  gravity. 

9.  The  truth  of  these  graphic  methods  may  be  shown  as  follows : 
Let  Oj  and  02,  Fig.  6,  be  the  positions  of  the  centers  of  gravity 

of  two  areas  A±  and  AZ9  respectively,  and  let  their  common  center 


of  gravity  be  situated  at  g,  distant  r±  from  0±  and  r2  from  02. 
By  the  principle  of  moments  we  shall  then  have  A^  =  A2r.2.  From 
02  lay  off  02b,  whose  length  represents  the  area  A^  to  some  selected 
scale,  and  from  0L  lay  off  O^d  parallel  to  02b  and  of  such  length  as 


272 


THE  ELEMENTS  OF  THE  MECHANICS  OF  MATERIALS 


to  represent  the  area  A2  to  the  same  scale.     Then  the  line  joining  b 
and  d  will  pass  through  g.    The  triangles  O±dg  and  02bg  are  similar, 


and  we  have  02b  :  O^d  =  02g  :  O^g,  or 


:  A2  =  r2 


whence 


A1rL  =  A2r2.     It  will  be  observed  that  the  lines  02b  and  O^d  are 
laid  off  on  opposite  sides  of  the  line  joining  Ot  and  02  and  at  oppo- 


site ends  to  their  respective  areas,  and  at  any  convenient  angle. 
Should  one  of  the  areas,  as  A2,  be  negative,  i.  e.,  is  the  area  of  a  hole, 
or  of  a  part  cut  out  of  the  surface,  then  02b  and  0±d  must  be  laid 
off  on  the  same  side  of  0±02,  as  shown  in  Fig.  7. 

10.  To  find  the  center  of  gravity  of  a  portion  of  a  regular  polygon 
or  of  an  arc  of  a  circle,  considered  as  a  thin  wire :   In  Fig.  8,  let 


riff.  8. 

L  =  length  of  the  sides  of  the  polygon,  or  of  the  length  of  the 
arc  in  the  case  of  a  circle;  R  =  radius  of  the  inscribed  circle,  or 
the  radius  of  the  circle  itself ;  C  =  chord  of  the  arc  of  .polygon  or 
circle;  Y"=  distance  of  the  center  of  gravity  from  the  center  of 

the  circle.     Then  L  :  E  —  C:  Y,  OT  Y  = -j-  .     For,  let  a  denote 


A 


the  length  of  a  side  of  the  polygon,  Fig.  9.  The  center  of  gravity 
of  each  side  will  be  at  its  middle  point,  and  distant  yl9  y2,  ys,  &c., 
from  the  diameter  of  the  inscribed  circle.  Let  x±,  x2,  XB,  &c.,  de- 


MOMENTS. — CENTER  OF  GRAVITY.  273 

note  the  projected  lengths  of  the  sides  on  the  diameter.     From  the 
similar   triangles   Ob c   and   edf  we   have   de  :  Ob  —  df  :  ~bc,   or 

a  :  R  —  xl  :  y1}  whence  y^  =  --~;  similarly,  y2  =  — ,  and  so  on. 

Cl  U, 

If  w  denotes  the  weight  of  each  of  the  n  sides  of  the  polygon,  we 
shal,  have :   T  =  %±$££  =  ?*±£-±**  =  *±*n±**. 

Substituting  the  values  of  yly  y2,  y3,  etc.,  we  have : 

-p 
y=  —  (x1  +  x2  -J-  &c.) .     But  na  =  L,  and  x±  +  xz  +  &c-  —  C> 

T>n 
hence  Y  =•  —j- .     The  polygon  becomes*  a  circle  when  n  is  infinitely 

great. 

11.  To  find  the  center  of  gravity  of  a  portion  of  a  regular  poly- 
gon or  of  a  sector  of  a  circle  when  considered  as  a  lamina  or  thin 
sheet. 


/o 


Keferring  to  Fig.  10,  and  using  the  same  notation  as  in  Art.  10, 

o  r> 
we  shall  have,  -o-  as  the  distance  of  the  center  of  gravity  of  each 

of  the  triangles  from  the  center  of  the  inscribed  circle,  and  they 
will  be  distant  ylf  yz,  yz,  &c.,  from  the  diameter.  The  base  of  each 
of  the  triangles  is  a,  and  the  projected  lengths  of  these  bases  on 
the  horizontal  diameter  are  xlf  x2,  xs,  &c.  From  the  similar  tri- 
angles 01  c  and  edf  we  have: 

2R  ZRx,  "2Rx.i 

--    :  a  =  y1  :  xlf  whence  y±  -    -        ;   similarly,  y2  -    -,  and 


so  on. 

If  w  denotes  the  weight  of  each  of  the  n  triangles  of  the  polygon, 
we  shall  have: 

w*  +  w*  +  &c-    i  +  «  +  &c-^R 


18 


274        THE  ELEMENTS  OF  THE  MECHANICS  OF  MATERIALS 

12.  To  show  the  application  of  the  foregoing  principles  to  finding 
centers  of  gravity,  the  solutions  of  a  few  problems  will  be  given. 

EXAMPLE  I. — Find  the  center  of  gravity  of  a  triangle. 

Conceive  the  triangle,  Fig.  11,  to  be  divided  into  a  great  number 
of  very  narrow  strips  drawn  parallel  to  one  of  the  sides,  as  B.  The 
center  of  gravity  of  each  strip  will  be  at  its  middle  point,  therefore 
the  center  of  gravity  of  the  triangle  will  lie  in  the  locus  of  these 
middle  points;  that  is,  in  the  median  Oc. 
The  elemental  area  of  the  triangle  =  b-dh. 


The  moment  of  the  elemental  area  —  h-b-dh. 

From  similar  triangles  we  have,  b  :  B  =  h  :  H ;  whence,  b  =  -„- 

Then, 

B-tf-dh 
Moment  of  elemental  area  =      —jj . 

p 

Moment  of  the  whole  area  =  T 

Ji 

ft  FJ 

The  area  of  the  triangle  •=.  — ~-  . 

The  distance,  x,  of  the  center  of  gravity  from  the  apex  is, 

BIT 

r  _  Moment  of  whole  area  _      3     _  2ff 
Whole  area  =  ~TT!l        3 

2 

That  is,  the  center  of  gravity  is  at  a  perpendicular  distance  below 
the  apex  equal  to  two-thirds  of  the  altitude,  and  must,  therefore, 
be  at  #,  the  intersection  of  the  median,  Oc,  and  the  parallel  to  the 

n  TT 

base  distant  -g-  from  the  apex.     From  similar  triangles  it  is  seen 

that  Og  is  f  the  length  of  the  median,  so  that  the  center  of  gravity 
is  on  the  median  at  two-thirds  its  length  from  the  apex. 


MOMENTS. — CENTER  OF  GRAVITY. 


275 


EXAMPLE  II. — A  square  is  divided  into  four  equal  triangles  by 
diagonals  intersecting  at  0;  if  one  triangle  be  removed,  find  the 
center  of  gravity  of  the  figure  formed  by  the  three  remaining  tri- 
angles. 

Let  w  denote  the  weight  of  each  of  the  remaining  triangles,  and 
let  a  denote  the  side  of  the  square,  Fig.  12.  The  weight  2w  of  the 


side  triangles  may  be  supposed  concentrated  at  their  common  center 
of  gravity,  0.     The  distance  of  the  center  of  gravity  of  the  lower 

triangle  from  0  is  ^  X  5  =  o  ,  by  Example  I.     Then  if  x  denotes 
the  distance  of  the  center  of  gravity  of  the  three  remaining  triangles 

Pp  +  Qq      2 to  X  0  -f  w  X  " 

from  0,  we  shall  have,  x  =  —  s ^  = s 

P  +  Q  2w  +  w 

That  is,  the  required  center  of  gravity  is  one-ninth  the  side  of  the 
square  from  the  center  of  the  square. 


a 

9* 


EXAMPLE  III. — A  quarter  of  the  area  of  a  triangle  is  cut  off  by 
a  line  drawn  parallel  to  the  base.  Find  the  center  of  gravity  of  the 
remaining  quadrilateral. 

Let  EF  be  parallel  to  the  base  5(7,  Fig.  13,  and  let  the  triangle 
AEF  be  the  part  cut  off.  The  required  center  of  gravity  will  lie 
in  the  median  AD.  The  triangles  AEF  and  ABC  are  similar,  and 


276        THE  ELEMENTS  OF  THE  MECHANICS  OF  MATEKIALS 


are  to  each  other  in  area  as  1  :  4.     Since  the  areas  of  similar  tri- 
angles are  to  each  other  as  the  squares  of  homologous  sides,  we  have  : 

_       _  ^2) 

1  :  4  =  AG2  :  AD2;    whence,  AG^—^-.     Let  x  denote  the  dis- 

tance of  the  required  center  of  gravity  of  the  quadrilateral  BEFC 
from  A.     Then, 


Pp-Qq 
~ 


4  y  %  X  AD      -       AD 
~~  ~" 


4-1  9 

That  is,  the  required  center  of  gravity  is  in  the  median  AD  and  at 
seven-ninths  of  its  length  from  A. 

EXAMPLE  IV.  —  Find  the  center  of  gravity  of  a  cone. 
Conceive  the  cone,  Fig.  14,  to  be  made  up  of  a  great  number  of 
thin  sections,  each  parallel  to  the  base.     The  center  of  gravity  of 


each  of  these  sections  will  be  at  its  center,  therefore  the  center 
of  gravity  of  the  cone  will  lie  in  the  locus  of  these  sections; 
that  is,  it  will  lie  in  the  line  joining  the  vertex  with  the  center  of 
gravity  of  the  base.  The  elemental  volume  =.  7rr~dh.  The  moment 
of  the  elemental  volume  =  irr^Tidli.  From  similar  triangles  we  have  : 

r  :  R  =  h  :  H  ;  whence,  r  =  --  .     By  substitution,  we  have  : 


Moment  of  elemental  volume  = 


™ 


7rR*   rHMl        *#2#2 

Moment  of  whole  volume  —  -gr  /     "<dti=   —  -  —  . 
Volume  of  cone  =      0     . 


MOMENTS.  —  CENTER  OF  GRAVITY.  277 

If  x  denotes  the  distance  of  the  center  of  gravity  of  the  cone  from 
the  vertex,  we  shall  have  : 


Moment  of  whole  volume  4          3H 


hole  volume 


3 

That  is,  the  center  of  gravity  is  at  a  perpendicular  distance  below 
the  apex  equal  to  three-fourths  of  the  altitude,  and  must,  therefore, 
be  at  g,  the  intersection  of  the  line  Oc  joining  the  apex  with  the 

O    TT 

center  of  gravity  of  the  base  and  the  parallel  to  the  base  distant  -j- 

from  the  apex.  From  similar  triangles  it  is  seen  that  Og  —  f  Oc,  so 
that  the  center  of  gravity  of  the  cone  is  in  the  line  joining  the 
vertex  with  the  center  of  the  base  and  at  three-fourths  its  length 
from  the  vertex. 


EXAMPLE  V.  —  Find  the  center  of  gravity  of  a  semicircular  arc, 
or  wire. 


In  Fig.  15  we  have,  from  Art.  10,  Y  =  ,  in  which  Y  =  dis- 
tance of  the  center  of  gravity  from  the  center,  L  =  length  of  the 
arc,  and  C  =  the  chord  of  the  arc.  Then,  Y  =  Rx£R  =^R  =  D 

~K  7T  7T* 

PKOBLEMS. 

1.  A  rod  3  feet  long  and  weighing  4  Ibs.  has  a  weight  of  2 
Ibs.  placed  at  one  end  ;  find  the  center  of  gravity  of  the  system. 

Ans.  One  foot  from  the  weighted  end. 

2.  Find  the  center  of  gravity  of  a  uniform  circular  disc  out  of 
which  another  circular  disc  has  been  cut,  the  latter  being  described 
on  a  radius  of  the  former  as  a  diameter. 

Ans.  One-sixth  of  radius  of  large  circle  from  center. 


278        THE  ELEMENTS  OF  THE  MECHANICS  OF  MATERIALS 

3.  A  heavy  bar  14  feet  long  is  bent  into  a  right  angle  so  that  the 
lengths  of  the  portions  which  meet  at  the  angle  are  8  feet  and  6  feet, 
respectively;   show  that  the  distance  of  the  center  of  gravity  of  the 
bar  so  bent  from  the  point  of  the  bar  which  was  the  center  of  gravity 

9  /2 

when  the  bar  was  straight,  is  -¥—  feet. 

4.  The  middle  points  of  two  adjacent  sides  of  a  square  are  joined 
and  the  triangle  formed  by  this  straight  line  and  the  edges  is  cut 
off;  find  the  center  of  gravity  of  the  remainder  of  the  square. 

Ans.  -^Y  of  diagonal  of  square  from  center  of  square. 

5.  A  piece  of  uniform  wire  is  bent  into  the  shape  of  an  isosceles 
triangle;   each  of  the  equal  sides  is  5  feet  long,  and  the  other  side 
is  8  feet  long ;  find  the  center  of  gravity. 

Ans.  -f%  °f  the  altitude  from  the  base. 

6.  Find  the  center  of  gravity  of  a  figure  consisting  of  an  equi- 
lateral triangle  and  a  square,  the  base  of  the  triangle  coinciding 
with  one  of  the  sides  of  the  square. 

Ans.  At  a  distance  from  the  base  of  the  triangle  equal  to 

J?    .  1        1  O    +    /C\A> 

of  the  base. 

7.  A  table  whose  top  is  in  the  form  of  a  right-angled  isosceles 
triangle,  the  equal  sides  of  which  are  three  feet  in  length,  is  sup- 
ported by  three  vertical  legs  placed  at  the  corners;   a  weight  of  20 
Ibs.  is  placed  on  the  table  at  a  point  distant  fifteen  inches  from 
oach  of  the  equal  sides ;   find  the  resultant  pressure  on  each  leg. 

Ans.  8J,  8J,  3J  Ibs. 

8.  ABCD  is  a  quadrilateral  figure  such  that  the  sides  AB,  AD, 
and  the  diagonal  AC  are  equal,  and  also  the  sides  CB  and  CD  are 
equal ;  find  its  center  of  gravity. 

Ans.  ??!+A*  units  from  C,  in  which  a  =  AB  and  I  =  CB. 
ba 

9.  ABC  represents  a  triangular  board  weighing  10  Ibs.     Suppose 
weights  of  5  Ibs.,  5  Ibs.,  and  10  Ibs.  are  placed  at  A,  B,  and  C,  respec- 
tively.    Where  is  the  center  of  gravity  of  the  whole? 

Ans.  At  five-ninths  of  the  median  drawn  from  C. 

10.  A  rod  of  uniform  thickness  is  made  up  of  equal  lengths  of 
three  substances,  the  densities  of  which  taken  in  order  are  in  the 


MOMENTS.  —  CENTER  OF  GRAVITY.  279, 

proportion  of  1,  2,  and  3  ;  find  the  position  of  the  center  of  gravity 
of  the  rod. 

Ans.  At  T7¥  of  the  whole  length  from  the  end  of  the  densest  part. 

11.  Find  the  position  of  the  center  of  gravity  of  a  piece  of  wire 
bent  to  form  three-fourths  of  the  circumference  of  the  circle  of 
radius  E. 

Ans.  On  a  line  drawn  from  the  center  of  the  circle  to  a  point 
bisecting  the  arc,  and  at  a  distance  0.3R  from  the 
center. 

12.  A  thin  wire  forms  an  arc  of  a  circle,  the  radius  of  which  is 
10  inches,  and  subtends  an  angle  of  60°.     Find  the  distance  of  the 
center  of  gravity  from  the  center.  Ans.  9.549  inches. 

13.  Find  the  position  of  the  center  of  gravity  of  a  balance  weight 
having  the  form  of  a  circular  sector  of  radius  E,  subtending  an 
angle  of  90°.  Ans.  0.6#  from  center  of  circle. 

14.  Find  the  center  of  gravity  of  a  parallelogram. 

Ans.  At  the  intersection  of  the  diagonals. 

15.  Find  the  center  of  gravity  of  a  semicircular  lamina,  or  sheet, 

9  7) 

of  diameter  D.  Ans.  ^  * 

O7T 

16.  A  square  stands  on  a  horizontal  plane;   if  equal  portions  be 
removed  from  two  opposite  corners  by  straight  lines  parallel  to  a 
diagonal,  find  the  least  portion  which  can  be  left  so  as  not  to  topple 
over.  4ns.  Three-quarters  of  the  area  of  the  square. 

17.  Find  the  center  of  gravity  of  a  trapezoid. 

Ans.  On  the  line  joining  the  middle  points  of  the  bases,  and 
at  a  perpendicular  distance  from  the  upper  base  equal 


to       •  -j      ,  and  from  the  lower  base, 


B  and  6  being  the  lower  and  upper  bases,  respectively, 
and  H  the  altitude.  If  the  distance  be  measured  on 
the  line,  8,  joining  the  middle  points  of  the  bases, 

S  H 

a  must  be  substituted  for  -^  • 

18.  Find  the  center  of  gravity  of  a  pyramid. 

.  Ans.  In  the  line  joining  the  vertex  with  the  center  of  gravity 
of  the  base  and  at  three-fourths  its  length  from  the 
vertex. 


,280        THE  ELEMENTS  OF  THE  MECHANICS  OF  MATERIALS 

19.  Find  the  center  of  gravity  of  a  frustum  of  a  cone. 

Ans.  In  the  line  joining  the  centers  of  gravity  of  the  upper 
and  lower  bases,  at  a  distance  from  the  upper  base 

,    ,     H    3^  +  2J5r  +  r2          ,    .          ,,      . 
equal  to   -^  •     pa   I    p     \     z~  >  and  from  the  lower 

H  3r2  +  2fir  +  #*    _, 

base,  -£-  •    jp  '     £r   '  ^  ,  R  and  r  being  the  radii  of 

the  lower  and  upper  bases,  respectively,  and  H  being 
the  altitude.  These  results  are  true  for  the  frustum 
of  any  pyramid  by  substituting  B  for  R  and  &  for  r, 
B  and  &  being  homologous  sides  of  the  lower  and  up- 
per bases,  respectively. 

20.  A  lever  safety-valve  is  required  to  blow  off  at  70  Ibs.  per 
square  inch.    Diameter  of  valve,  3  inches ;  weight  of  valve,  3  Ibs. ; 
short  arm  of  lever,  2.5  inches;  weight  of  lever,  11  Ibs.;  distance  of 
center  of  gravity  of  lever  from  fulcrum,  15  inches.     Find  the  dis- 
tance at  which  a  cast-iron  ball  6  inches  in  diameter  must  be  placed 
from  the  fulcrum.    The  weight  of  the  ball-hook  =  0.6  Ib. 

Ans.  35.48  in. 

21.  A  steel  safety-valve  lever  1  inch  thick  and  50  inches  long 
tapers  from  3  inches  in  depth  at  the  fulcrum  to  1  inch  at  the  end. 
It  overhangs  the  fulcrum  4  inches,  the  overhang  having  no  taper. 
Diameter  of  valve,  4  inches;   weight  of  valve,  4.75  pounds;   short 
arm  of  the  lever,  3.5  inches.     Find  the  distance  from  the  fulcrum 
at  which  a  cast-iron  ball  9.5  inches  in  diameter  must  be  placed  in 
order  that  steam  shall  blow  off  at  125  Ibs.  pressure  per  square  inch. 
Weight  of  the  ball-hook,  1.3  Ibs.;   weight  of  a  cubic  inch  of  steel, 
0.28  Ib. ;  weight  of  a  cubic  inch  of  cast  iron,  0.26  Ib. 

Ans.  42.35  inches. 

22.  A  trapezoidal  wall  has  a  vertical  back  and  a  sloping  front 
face;  width  of  base,  10  ft. ;  width  of  top,  7  ft. ;  height,  30  ft.     What 
horizontal  force  must  be  applied  at  a  point  20  ft.  from  the  top  in 
order  to  overturn  it,  i.  e.,  to  make  it  pivot  about  the  toe  ?    Width 
of  wall,  1  ft.;  weight  of  masonry  in  wall,  130  Ibs.  per  cubic  foot. 

Ans.  18,900  Ibs. 

23.  Find  the  height  of  the  center  of  gravity  from  the  base  of  a 
column  4  feet  square  and  40  feet  high,  resting  on  a  tapered  base 
forming  a  frustum  of  a  square-based  pyramid  10  feet  high  and 
8  feet  square  at  the  base.  Ans.  20.4  feet  from  base. 


MOMENTS. — CENTER  OF  GRAVITY.  281 

Problems  24,  25,  26,  and  27  are  to  be  solved  graphically. 

24.  Find  the  position  of  the  center  of  gravity  of  an  unequally 
flanged  beam-section;   top  flange,  3  inches  wide,  1.5  inches  thick; 
bottom  flange,  15  inches  wide,  1.75  inches  thick;  web,  1.5  inches 
thick;  total  height,  18  inches. 

Ans.  5.72  inches  from  the  bottom  edge. 

25.  Find  the  height  of  the  center  of  gravity  of  a  T  section  from 
the  foot,  the  top  cross-piece  being  12  inches  wide  and  4  inches 
deep ;  the  stem,  3  feet  deep  and  3  inches  wide. 

Ans.  24.16  inches.  (Check  the  result  by  seeing  if  the  mo- 
ments are  equal  about  a  line  passing  through  the  sec- 
tion at  the  height  found. ) 

26.  A  square  board  weighs  4  Ibs.,  and  a  weight  of  2  Ibs.  is  placed 
at  one  of  its  corners.     Find  the  position  of  the  center  of  gravity  of 
the  board  and  weight. 

Ans.  On  the  diagonal  drawn  from  the  weighted  corner  and 
at  two-thirds  its  length  from  the  opposite  corner. 

27.  ABCD  is  a  flat  plate,  right-angled  at  5,  and  having  the  fol- 
lowing dimensions :  AB  =  15  inches;  AD  =  12.5  inches,  EG  =  18 
inches,  and  the  perpendicular  from  D  on  EG  measures  18.5  inches. 
Construct  the  figure  to  a  scale  of  1.5  inches  to  the  foot,  and  locate 
the  center  of  gravity  of  the  plate,  measuring  its  distance  from  the 
side  AB.  Ans.  7.895  inches. 


CHAPTER  II. 


BENDING  MOMENT. 
DIAGRAM. 


SHEAR.     BENDING-MOMENT 
SHEAR  DIAGRAM. 


13.  Bending  Moments. — When  forces  act  on  a  body  in  such  a 
manner  as  to  tend  to  give  it  a  spin  or  a  rotation  about  an  axis  with- 
out any  tendency  to  shift  its  center  of  gravity,  the  body  is  said  to 
be  acted  on  by  a  couple.  A  couple  consists  of  two  parallel  forces 


of  equal  magnitude  acting  in  opposite  directions,  but  not  in  the 
same  straight  line. 

A  beam  is  subjected  to  a  bending  moment  when  it  is  so  acted  upon 
at  its  ends  by  equal  and  opposite  couples  that  there  is  a  tendency 
to  turn  it  in  opposite  directions. 

Thus  the  beam  ran,  Fig.  16,  is  acted  on  by  the  equal  and  oppo- 
site couples,  E  and  R,  and  TF  and  IF,  the  tendency  being  to  turn 


* 


1 


the  beam  in  opposite  directions  about  the  point  r;  that  is,  to  bend 
it  at  r.  In  Fig.  17  the  couples  whose  moments  are  R^  X  mr  and 
R2  X  nr  have  the  same  effect  on  the  beam  as  those  of  Fig.  16.  The 
beam  of  Fig.  16  is  called  a  cantilever,  from  the  nature  of  its  sup- 
port, while  the  beam  of  Fig.  17  is  called  a  simple  beam. 

14.  General  Case  of  Bending  Moments. — The  bending  moment  at 
any  section  of  a  beam  is  the  algebraic  sum  of  all  the  moments  of 


BENDING  MOMENT. — SHEAR.  283 

the  external  forces  acting  to  the  left  of  the  section.  It  is  equal  to 
the  moment  of  the  reaction  at  the  left  support  minus  the  sum  of 
the  moments  of  the  loads  to  the  left  of  the  section  considered.  It  is 
thus  assumed  that  forces  acting  upward  are  positive,  and  those 
acting  downward  are  negative,  so  that  bending  moments  may  be  posi- 
tive or  negative,  depending  upon  whether  the  moment  of  the  left 
reaction  is  greater  or  less  than  the  sum  of  the  moments  of  the  loads 
to  the  left  of  the  section. 

15.  In  graphical  constructions  the  signs  of  bending  moments  are 
cf  the  first  importance  and  are  determined  by  the  following  rule : 

Bending  moments  which  tend  to  bend  a  beam  or  cantilever  con- 
cave upward, ,  are  regarded  as  positive,  and  when  they  tend 

to  bend  in  the  reverse  way,  , — •% ,  they  are  negative. 

16.  It  should  be  observed  that  it  is  merely  to  avoid  confusion  in 
the  construction  of  diagrams  that  the  external  forces  to  the  left  of 
the  section  were  considered  when  defining  the  bending  moment  at 
any  section  of  a  beam.     Moments  may  equally  well  be  taken  to  the 
right  of  the  section  and  the  same  value  be  obtained  for  the  bending 
moment  at  the  section.     When  bending  moments  are  obtained  by 
calculation,  rather  than  by  construction,  the  side  involving  the  least 
calculation  in  taking  moments  should  be  selected,  though  the  cal- 
culations for  both  sides  afford  a  positive  check  as  to  the  accuracy  of 
the  work. 

17.  The  beam,  Fig.  18,  is  supposed  to  be  without  weight.     The 
bending  moment  at  the  section  E  is,  taking  moments  to  the  left  of  E, 

M^R^XaE  —  WiXqE  —  WzX  pE,  (1) 

or,  taking  moments  to  the  right  of  E,  we  have : 

M  =  R2XbE  —  W3XrE.  (2) 

Suppqse  the  distances  and  weights  to  be  as  shown  in  the  figure. 
Before  finding  the  bending  moments  we  must  first  find  the  reactions, 
R!  and  R2. 

Taking  moments  about  a,  we  have : 

R2  X  10.5  =  80  X  8.5  +  50  X  5  +  60  X  2,  whence 

R2  =  100  Ibs. 
Taking  moments  about  &,  we  have : 

#±  X  10.5  =  60  X  8.5  +  50  X  5.5  +  80  X  2,  whence 

R±  =  90  Ibs. 
which   might   have   been   expected,    since   Rl  -f-  R0    should    equal 


284        THE  ELEMENTS  OF  THE  MECHANICS  or  MATERIALS 

By  substitution  in  equation  (1),  we  have  for  the  bending  mo- 
ment at  E, 

M,  =  90  X  6.5  —  60  X  4.5  —  50  X  1.5  =.  240  pounds-feet. 
Substituting  in  equation  (2),  we  have: 

M  —  100  X  4  —  80  X  2  =  240  pounds-feet. 

It  is  thus  seen  that  the  bending  moment  at  a  section  is  the  same, 
whether  the  moments  be  taken  to  the  left  or  to  the  right.  In  this 
instance  the  taking  of  moments  to  the  right  involved  the  least  cal- 
culation. 


18.  Bending-moment  Diagrams  are  made  to  show  graphically  the 
bending  moment  at  any  section  of  a  beam. 

For  example,  take  the  beam  of  Fig.  18.  The  bending  moment, 
Jf,  at  q  is  M  —  R^  X  2  =  90  X  2  =  180  Ibs.-ft. 

At  p,  M  —  R^  X  5  —  Ft  X  3  =  90  X  5  —  60  X  3  =  450  —  180 
=  270  Ibs.-ft. 

At  r,  M  =  R^X  8.5  —  W ,  X  6.5  —  W2  X  3.5 

=  90  X  8.5  —  60  X  6.5  —  50  X  3.5  —  200  Ibs.-ft. 

If,  on  a  base  line  a'b',  Fig.  19,  and  to  a  scale  of  1  inch  =  200 
pounds-feet,  we  erect  ordinates  to  represent  these  bending  moments, 
and  then  join  their  extremities  with  the  broken  line  a'q'p'r'V,  the 


BENDING  MOMENT. — SHEAR. 


285 


inclosed  figure  is  a  diagram  whose  ordinate  beneath  any  section  of 
the  beam  will  measure  the  bending  moment  at  the  section  to  the 
scale  adopted.  Thus  the  ordinate  for  the  bending  moment  at  q 

-i  o/-v  2*70 

measures  <^  =  0.9  inch;    that  at  p,  ^^  =  1.35  inch;   that  at  r, 

200 
200 


=  1  inch.     The  ordinate  beneath  E  measures  1.2  inch,  which, 


to  scale,  represents  a  bending  moment  of  1.2  X  200  =  240  pounds- 
feet,  as  already  found  by  calculation. 

19.  Shear. — Shearing  stresses  exist  when  couples,  acting  like  a 
pair  of  shears,  tend  to  cut  a  body  between  them.  The  application 
of  couples  to  beams  in  the  manner  already  described  subject  the 
beams  to  shearing  actions  as  well  as  to  bending  moments.  With 
beams  of  lengths  ordinarily  encountered,  the  bending  moment  is  far 


more  important  than  the  shear,  but  with  very  short  simple  beams 
and  very  short  cantilevers,  failure  will  invariably  result  from  shear. 

20.  Shear  Diagrams  are  made  to  show  the  amount  of  the  shear  at 
any  section  of  a  beam,  and  in  their  construction  attention  must  be 
paid  to  the  signs,  so  that  in  cases  where  the  shear  is  partly  positive 
and  partly  negative  the  positive  part  may  be  placed  above  the  shear 
axis,  or  base  line,  and  the  negative  part  below. 

21.  General  Case  of  Shear. — The  shear  at  any  section  of  a  beam 
or  of  a  cantilever  is  equal  to  the  left  reaction  minus  the  sum  of  the 
loads  between  the  reaction  and  the  section.     In  other  words,  it  is 
the  algebraic  sum  of  the  forces  to  the  left  of  the  section,  the  upward 
forces  being  regarded  as  positive  and  the  downward  negative. 

22.  If  the  short  cantilever,  Fig.  20,  be  loaded  until  it  fails,  there 
will  be  a  slight  bending  at  the  outer  end,  but  the  failure  will  be 
due  to  the  outer  part  shearing  off  bodily  from  the  built-in  part,  as 


286 


THE  ELEMENTS  OF  THE  MECHANICS  OF  MATEBIALS 


indicated  by  the  dotted  lines.  As  there  is  no  reaction  at  the  left 
end  of  the  beam,  the  shear  at  all  sections  is  constant  and  equal  to 
-  W.  The  sign  of  the  shear  may  be  determined  by  a  consideration 
of  the  sliding  tendency  of  the  shear — whether  clockwise  or  contra- 
clockwise.  In  Fig.  20,  the  sliding  tendency  is  contra-clockwise,  and 
therefore  negative.  The  ordinates  of  the  shaded  shear  diagram  are 
of  constant  length  and  equal  to  —  W,  and  the  whole  of  the  dia- 
gram, being  negative,  is  below  the  shear  axis  mn. 

23.  In  the  case  of  the  simple  beam,  Fig.  21,  a  failure  from  shear 
will  occasion  the  part  between  the  supports  to  slide  down,  as  shown 
by  the  dotted  lines.  The  shear  at  any  section  to  the  left  of  the 
load,  W,  is  R1}  and  at  any  section  between  W  and  the  right  support 


it  is  R1  —  W  =  —  R2,  since  R1  +  R2  —  W.  The  clockwise  ten- 
dency of  the  slide  of  the  shear  at  the  left  support,  and  the  contra- 
clockwise  tendency  at  the  right  support  determine  the  signs  of  the 
shears  as  shown  by  the  arrows. 

The  shaded  part  of  Fig.  21  represents  the  shear,  and  shows  that  it 
changed  sign  under  the  load,  the  positive  part  being  placed  above 
the  shear  axis,  mn,  and  the  negative  part  below. 

24.  As  in  the  case  when  denning  bending  moments,  only  the 
external  forces  to  the  left  of  a  section  were  considered  when  defining 
the  shear  at  the  section,  but  this,  as  in  the  case  of  bending  moments, 
was  only  for  convenience.  If  W8  denotes  the  sum  of  all  the  loads 
between  the  left  support  and  a  given  section,  and  W8'  the  sum  of 
all  the  loads  between  the  section  and  the  right  support,  then  evi- 
dently R!  +  R2  —  W8  +  W9',  whence  7^  —  W8  —  —  (R2  —  Ws'). 
The  first  member  of  this  equation  is  the  algebraic  sum  of  the  forces 


BENDING  MOMENT. — SHEAR.  28? 

to  the  left  of  the  given  section  and  is  the  shear  at  the  section,  and 
the  second  member  is  the  negative  of  the  algebraic  sum  of  the  forces 
to  the  right  of  the  section.  The  shear  at  a  given  section  is  then 
the  algebraic  sum  of  the  external  forces  to  the  right  or  to  the  left  of 
the  section,  but  with  contrary  signs. 

25.  Relations  between  Bending-moment  and  Shear  Diagrams. — 
The  depth  of  the  bending-moment  diagram  beneath  any  section 
measures,  as  already  shown,  the  bending  moment  at  the  section  to  a 
given  scale,  and  is  equal  to  the  algebraic  sum  of  the  moments  of  the 
external  forces  to  the  left  of  the  section. 

The  depth  of  the  shear  diagram  beneath  any  section  shows,  to  a 
given  scale,  the  shear  at  the  section,  and  is  equal  to  the  algebraic 
sum  of  the  external  forces  to  the  left  of  the  section. 

Let  the  beam,  Fig.  22,  be  loaded  with  W19  W2,  W3,  at  distances 
d19  d2,  d3,  respectively,  from  any  given  section,  a.  Denote  the  sup- 
port reactions  by  R^  and  R2,  and  their  distances  from  a  by  r±  and  r2, 
respectively. 

The  shear  at  the  extreme  right  end  is,  by  our  definition, 
Ri  —  Wi  —  W2  —  WQ  —  —  R2.  On  a  base  line,  mn,  Fig.  23,  con- 
struct separately  the  component  parts  of  this  shear,  giving  to  .each 
component  part  its  algebraic  sign.  Thus  the  rectangle  mq  repre- 
sents the  shear  throughout  the  beam  due  to  R^ ;  in  like  manner  the 
rectangles  qs,  tu,  and  uv,  represent,  respectively,  the  shears  due  to 
TFj,  W2,  and  W3,  and  the  whole  shaded  figure  is  the  shear  diagram 
of  the  beam,  the  algebraic  sum  of  the  shears  beneath  any  section  of 
the  beam  measuring  the  shear  at  the  section  to  the  scale  of  the 
figure. 

The  area  of  this  shear  diagram  to  the  left  of  the  section  a  is  equal 
to  R^  —  W^.  (1) 

The  area  to  the  right  of  section  a  is  equal  to 
Rir2-Ws2-W2(r2-d2)-Ws(r2-d3) 
=  R±r2  —  TP±r0  —  F,r0  +  W0d2  —  Wsr2  +  W3d3 
=  r2(R,  —  W~—  W2  —  W3)  +  W2d2  +  W3d3 
=—R,rt  +  W&  +  W9d9 
=  -(R2r2-W2d2-^W3d5).'  (2) 

The  second  member  of  equation  (1)  is,  by  our  definition,  the 
bending  moment  to  the  left  of  section  a,  and  the  second  member  of 
equation  (2)  is  the  bending  moment  to  the  right  of  section  a,  and 
since  the  bending  moment  at  any  section  is  the  same,  whether  taken 


288        THE  ELEMENTS  OP  THE  MECHANICS  OF  MATEEIALS 

to  the  right  or  to  the  .left  of  the  section,  it  follows  that  the  area  of 
the  shear  diagram  to  the  right  or  to  the  left  of  a  section  is  equal  to 
the  bending  moment  at  the  section,  having  regard  always  to  the 
algebraic  signs. 

26.  Figure  23  is  an  inconvenient  form  of  the  shear  diagram.     If 
the  algebraic  sum  of  the  shears  be  taken  at  every  section  of  the  beam, 
and  all  those  that  are  plus  be  plotted  above  the  shear  axis  mn,  and 
those  that  are  minus  below,  the  ordinate  of  the  resulting  diagram 
beneath  any  section  will  be  the  measure  of  the  shear.     The  easiest 
way  of  finding  these  algebraic  sums  is  to  revolve  the  negative  part 
of  Fig.  23  about  pq,  the  upper  boundary  of  the  positive  part,  so 
that  it  assumes  the  position  indicated  by  the  dotted  lines.     By  this 
process  of  superposition  the  major  part  of  the  positive  shear  is  neu- 
tralized, and  the  resulting  diagram  is  that  shown  in  Fig.  24.     Fig. 
23  is  a  cumbersome  way  of  obtaining  the  shear  diagram,  and  was 
constructed  only  to  demonstrate  the  relations  between  the  shear  and 
bending-moment  diagrams.     The  proper   procedure  would  be  to 
derive  Fig.  24  at  once  from  Fig.  22.     Thus,  Rlf  W19  W2,  and  W3 
having  been  determined  to  scale,  and  the  shear  axis,  or  base  line, 
m'n',  selected,  we  would  proceed  as  follows  : 

The  shear  between  the  left  support  and  TF±  is  equal  to  R19  which 
gives  the  point  &.  Immediately  the  point  of  'application  of  W^  is 
passed  the  shear  becomes  R^  —  W19  giving  the  point  c,  and  this 
shear  continues  up  to  the  point  of  application  of  W2,  but  imme- 
diately this  point  is  passed  the  shear  becomes  Ri  —  Wt  —  W2,  giv- 
ing the  point  d.  This  shear  continues  until  the  point  of  application 
of  Ws  is  passed  when  the  shear  becomes  R^  —  TF±  —  W2  —  W3 
=  —  R2,  giving  the  point  e. 

27.  Figures  22,  23,  and  24  were  constructed  to  the  following 
scales  : 

Length,  1  inch  =  4  feet;  load,  1  inch  =  150  IDS.;  so  that  1  sq. 
inch  =  150  X  4  =  600  Ibs.-ft. 

The  data  of  the  beam  and  loads  are  as  follows  : 

Length  of  beam  =  12  ft.  =  -1/  =  3  inches  to  scale;    W^  =  60 

Ibs.  =        =  0.4  inch  to  scale;  W2  =  45  Ibs.  =         =  0.3  inch; 


QO 

W  =  90  Ibs.  =  ^  =  0.6  inch;  ^  =  2  f  t.  =  f  =  -J  inch;  d2  =  3 
ft.  =  }  =  0.75  inch;  dz  =  6  f  t.  =  £  =  1.5  inch;  r±  =  5  ft  =  J 
=  1.25  inch;  r,  =  7  ft.  =  J  =  1.75  inch. 


r 


cc, 


m$ 


i  T 


1 

"w  A* 

-f 

B^* 

\ 

z 

!      I 

1 

H 

f 

Ml 

^        Fig-Z-^ 

. 

1 

ill 

c 

-> 

b,  «  , 

17? 

Tjl-                                       1    ' 

jf                                  — 

^                                        II 

II 

d,          T 

pi         * 

Fiq.  Z5^           6 

ii  in 

1 


290        THE  ELEMENTS  OF  THE  MECHANICS  OF  MATERIALS 

The  support  reactions  were  first  found  as  follows: 
Taking  moments  about  the  left  support,  we  have : 

*.(»-i  +  »•«)  =  IF. ('i  +  *,)  +  ^,('1  +^)  +  ^(r,  -  dj,  or 
=  90  X  11  +  45  X  8  +  60  X  3,  whence  B2  =  127.5  Ibs. 

K 

=  0.85  inch.    Therefore,  R±  =  60  +  45  +  90  —  127.5 
=  67.5  Ibs.  =  ~  —  0.45  inch. 

.  10U 

28.  The  shear  diagram,  Fig.  24,  not  only  gives  by  its  ordinate 
under  any  section  the  shear  at  that  section,  but  it  also  gives  the  bend- 
ing moment  at  any  section  by  taking  the  algebraic  sum  of  the  areas 
to  the  right  or  to  the  left  of  the  section.     It  follows  from  this  that 
the  positive  area  of  the  diagram,  or  the  area  above  the  base  line,  is 
numerically  equal  to  the  negative  area,  or  the  area  below  the  base 
line. 

The  algebraic  sum  of  the  areas  to  the  right  or  to  the  left  of  sec- 
tion a  is  found  by  measurement  to  be  0.3625  sq.  inch;  the  bending 
moment  at  that  section  is,  therefore,  0.3625  X  600  =  217.5  Ibs.-ft. 

29.  The  bending-moment  diagram,  Fig.  25,  was  constructed  on 
the  base  line  m"n"  to  a  scale  such  that  1  inch  in  depth  =  300  Ibs.-ft. 
The  bending  moments  at  Wly  W2,  W3  were  calculated  as  follows: 

MWi  =  RI(TI  —  dj  =  67.5  X  3  =  202.5  Ibs.-ft.  =  ^5  _  0.675 

inch  to  scale. 
M9  =  R^fa  +  d2)  -  -  W1  X  d2  =  67.5  X  8  —  60  X  5  =  240 

240 
pounds-feet  =  O?TQ  =  0.8  inch. 

*1  =  £,(*!  +  d3)  -  T^K  +  dt)  -  W,(ds  -  d.) 

1  27  5 

=  67.5  X  11  —  60  X  8  —  45  X  3  =  127.5  Ibs.-ft.  =    ™£ 

oUU 

=  0.425  in. 

Ordinates  equal  in  height  to  the  scale  measurements  representing 
these  bending  moments  were  erected  on  m"n",  as  shown,  and  their 
extremities  joined  by  the  broken  line  m'Jghlcn".  The  ordinate  of 
the  resulting  diagram  beneath  any  section  of  the  beam  is  a  scale 
measurement  of  the  bending  moment  at  the  section.  Thus  the  or- 
dinate under  the  section  a  measures  |$  of  an  inch,  corresponding  to 
a  bending  moment  of  |-§  X  300  =  217.5  Ibs.-ft.,  as  already  found 
from  the  shear  diagram. 


BENDING  MOMENT. — SHEAR. 


291 


30.  Figure  26  represents  a  simple  beam  supported  at  the  ends  and 
uniformly  loaded  with  w  pounds  per  unit  of  length.  The  total  load 
is  wL  Ibs.,  and  it  is  evident  that  the  support  reactions  will  each 


^um» 

o  '' 


wL 


equal  -^-  Ibs.  To  find  the  bending  moment  at  any  section,  a,  dis- 
tant x  from  the  left  support,  we  may  assume  the  uniform  load  to  be 
made  up  of  a  number  of  parallel  forces  each  equal  to  w.  There  are 
wx  of  these  forces  between  section  a  and  the  left  support,  and  we 


292        THE  ELEMENTS  OF  THE  MECHANICS  OF  MATERIALS 

may  substitute  for  them  their  resultant,  wx,  acting  at  their  center 
of  gravity,  which  is  midway  between  the  section  and  the  support. 
The  bending  moment,  M  a,  at  the  section  is  then 

•M-        T>  ..  x      wLx      wx*      wx  ,  T         -.       wxx' 

Ma  =  R,x  —  wx  X     =  --  -  --  =  •-  (L  —  x)  = 


That  is,  the  bending  moment  at  the  section  is  proportional  to  the 
product  of  the  segments  into  which  the  section  divides  the  beam, 
and  the  bending-moment  diagram  is,  therefore,  a  parabola,  as  shown 
in  Fig.  27,  having  its  axis  vertical  and  under  the  middle  of  the  beam. 
For  it  is  a  property  of  the  parabola  that,  if  a  diameter  be  drawn  to 
intersect  a  chord,  then  the  product  of  the  segments  of  the  chord  is 
proportional  to  the  length  of  that  part  of  the  diameter  included 
between  its  vertex  and  its  point  of  intersection  with  the  chord. 
If,  in  the  general  expression  for  the  bending  moment, 

•a*         wLx       WX*          i  ,  L  ,,      ., 

Ma  =  —  £—  —  -£-  ,  we  let  x  =  ^  ,  we  shall  have  for  the  maximum 

bending  moment,  Mmax, 

_w!S_tj>IS  =  wL*  =  WL 

Mmax  -  •     4  y  g  y     , 

in  which  W  =  wL  =  the  whole  weight. 

31.  The  shear  diagram,  Fig.  28,  was  constructed  as  follows: 
The  shear  at  the  extreme  right  end  is  R1  —  wL  =.  —  R2,  or 

~2  --  wL  =  —  -£-.  Constructing  the  component  parts  of  this 
shear  we  get  the  rectangle  mg  as  the  shear  throughout  the  beam  due 
to  —•  .  Conceiving  the  uniformly  distributed  load  to  be  made  up 

of  L  units  of  w  Ibs.  each,  the  shears  of  these  units  will  be  repre- 
sented by  the  horizontal  lines  of  the  triangular  part  of  the  diagram, 
and  are  negative.  Eevolving  the  negative  part  of  the  diagram  about 
rg,  the  upper  boundary  of  the  positive  part,  until  it  assumes  the 
position  indicated  by  the  dotted  lines,  we  get  the  resulting  shear 
diagram  of  Fig.  29. 

32.  Figure  29  may  readily  be  obtained  at  once  from  Fig.  26. 
Thus  the  shear  at  any  section  distant  £  from  the  left  support  is 

Y  =  w  ----  wx,  Y  denoting  the  shear.     This  is  the  equation  of  a 

straight  line,  the  origin  being  at  the  left  support.  It  is  seen  that 
the  shear  decreases  as  x  increases,  and  has  its  maximum  value. 

^5-,  when  x  —  0.     When  x  =   ~  the  value  of  Y  becomes  0,  and 


BENDING  MOMENT. — SHEAR.  293 

when  x  is  greater  than  -~  the  value  of  F  becomes  negative,  and 
when  x  =  L  the  value  of  F  becomes  —  -^- .  r's"  is  then  the 
straight  line  whose  equation  is  F  =  -~ wx. 

33.    Bending-moment    and   Shear   Diagrams    of    Cantilevers. — 

Figure  30  represents  a  cantilever  with  a  concentrated  load,  W,  at  the 
end,  a  concentrated  load,  TP±,  at  an  intermediate  point  between  the 
end  and  the  support,  and  a  uniformly  distributed  load  of  w  pounds 
per  foot  over  a  portion  of  its  length.  The  construction  of  the  bend- 
ing-moment  and  shear  diagrams  of  this  beam  will  serve  to  illustrate 
the  three  cases  of  a  cantilever:  (a)  Loaded  at  the  end.  (&)  A  con- 
centrated load  at  some  point  between  the  end  and  the  support,  (c) 
Loaded  uniformly  with  w  pounds  per  lineal  foot. 

The  cantilever  of  Fig.  30  has  been  drawn  so  that  there  is  no  reac- 
tion at  the  left  end,  as  can  be  done  in  all  cases.  From  our  defini- 
tion of  bending  moment  it  will  be  seen  then  that  the  bending 
moment  is  zero  at  the  free  end  and  is  a  maximum  at  the  wall.  The 
tendency  being  to  bend  the  beam  concave  downward  all  the  bending 
moments  are  negative. 

The  bending  moment  at  any  point,  x,  between  W  and  W±  is 
M  =  —  Wx.  At  any  point,  xlf  between  W^  and  the  point  of  com- 
mencement of  the  uniform  load  of  w  Ibs.  per  foot  is  M  =  —  Wx± 
—  ^1(^1  —  a)>  a  being  the  distance  between  W  and  W±.  At  any 
point,  ar2,  the  bending  moment  is 

M=  —  Wx2  —  Wi(xt  —  a)  —  w  \  Xt  —  (a  -j- 


the  factor  X<L  ~  ^?      — -  being  the  arm  of  the  center  of  gravity  of 

that  portion  of  the  beam  that  is  uniformly  loaded  included  by  the 
abscissa  xz.     The  maximum  bending  moment  at  the  wall  is 

Mmax  =  —  WL  —  Wsi  —  wr2  X  \  =  —  WL  —  W,r,  —  ^  . 

34.  In  the  construction  of  the  bending-moment  diagram,  Fig.  31, 
the  following  data  were  used : 
Lineal  scale,  0.2  inch  =  1  foot. 
Load  scale,  1  inch  in  depth  =  800  Ibs.  ft. 


294        THE  ELEMENTS  OF  THE  MECHANICS  OF  MATERIALS 

L  =  10  ft.  =  0.2  X  10  —  2  inches  to  scale;  rx  =  7  ft.  =  0.2  X  7 
=  1.4  inches;  r2  =  4  ft.  =  0.2  X  4  =  0.8  inch;  F  =  32  Ibs.  ; 
F±  z=  44  Ibs.  ;  w  =  48  Ibs.  per  foot  run.  Then,  the  bending  mo- 
ment at  the  wall  is  : 


Mmax  =  -  -     32   X  10  +  44  X  ?  +  -_       -=  _  1012  Ibs.-ft. 


inches  to  scale,  =  ordinate  of  bending-mo- 
ment diagram  at  the  wall. 

MWi  =  —  W(L  —  rj  =  —  32  X  3  =  —  96  lbs.-ft.  =  — ^ 

=  —  0.12  inch. 

ME  =  W(L  —  r2)  -  -  Fifo  --  r2)   =  --  32  x  6  —  44  X  3 
=  —  324  lbs.-ft.  =  —  ??*  =  —  0.405  inch. 

oUU 

Erecting  these  three  ordinates  the  bending-moment  diagram  is  com- 
pleted, the  curve  under  the  part  of  the  beam  that  is  uniformly  loaded 
being  parabolic. 

The  ordinates  corresponding  to  the  bending  moments  at  W j.  and 
at  E  might  have  been  found  graphically  as  follows : 

Divide  the  maximum  bending  moment  (represented  by  the  or- 
dinate at  the  fixed  end  of  the  beam)  into  its  component  parts,  WL, 

F^,  and   ^p,  equal  respectively  to  320,  308,  and  384  lbs.-ft.,  and 

equal,  when  reduced  to  scale,  to  0.4  inch,  0.385  inch,  and  0.48  inch, 
respectively.  The  ordinates  for  the  bending  moments  at  F T  and  at 
E  are  then  found  as  shown  by  the  dotted  line  of  the  figure. 

35.  In  the  construction  of  the  shear  diagram,  Fig.  32,  the  lineal 
scale  of  the  bending-moment  diagram,  J  inch  =  1  foot,  was  adopted 
but  the  load  scale  was  taken  as  1  inch  =  160  Ibs.,  so  that  1  sq.  in. 
of  area  of  the  shear  diagram  —  160  X  4  =  640  lbs.-ft. 

Since  the  sliding  tendency  is  contra-clockwise  all  the  shear  is 
negative,  and  is  a  maximum  at  the  wall.  Since  there  is  no  reaction 
at  the  left  end  the  maximum  shear  at  the  wall  =  —  W  —  F±  —  wr2 

=  —  32  —  44  —  48  X  4  =  —  268  Ibs.  =  —  ^  =  —  1.675  inch 

to  scale.  The  component  parts  of  this  shear  are  the  shears  due  to 
F,  W19  and  wr2,  viz.,  32  Ibs.,  44  Ibs.,  and  192  Ibs.,  equal,  respectively, 
to  0.2  inch,  0.275  inch,  and  1.2  inch,  when  reduced  to  scale.  Con- 
struct these  component  shears  as  shown  in  the  figure,  and  the  shear 


BENDING  MOMENT. — SHEAR. 


295 


diagram  is  complete.  The  are^  of  this  diagram  to  the  left  of  any 
section  is  equal  to  the  bending  moment  at  the  section.  Thus,  the 
area  to  the  left  of  E  is  found  by  measurement  to  be  0.50625  sq.  inch, 
which,  when  reduced  to  scale,  equals  640  X  0.50625  =  324  lbs.-ft, 
the  same  as  was  found  above  by  calculation. 


36.  The  beam  of  Fig.  33  is  supported  at  the  ends  and  has  a  cen- 
tral load  W. 

W 

The  reaction  at  each  support  is  -^  .     The  bending  moment  at  any 

section,  x,  between  the  left  support  and  the  middle  is  Mx  =  —9— » 


296        THE  ELEMENTS  OF  THE  MECHANICS  OF  MATERIALS 


and  increases  directly  as  x  increases.     At  the  middle  x  =  ^  ,  and 
M  =  —  -.     At  any  section,  x±,  between  the  middle  and  the  right 


8upport,  MXl  =         -  W      -=- 


= 


and  decreases  as  x±  increases,  becoming  0  when  xt  •=.  L.     The  maxi- 
mum bending  moment  is,  therefore,  at  the  middle  and  is  equal  to 


.    Since  Mx  —       and  MX1  =     (L  —  ar±)  are  straight  line  equa- 


tions,  the  bending-moment  diagram  has  the  triangular  form  shown 
in  Fig.  34. 

W 

37.  The  shear  equals  •«-  until  directly  after  passing  the  middle 

W  W 

section,  when  it  becomes  -«-  —  W  =  --  g-  .     The  shear,  therefore, 

changes  sign  at  the  middle.    The  shear  diagram  is  then  as  shown  in 
Fig.  35. 


BENDING  MOMENT. — SHEAR. 


297 


38.  The  beam  of  Fig.  36  is  supported  at  the  ends  and  has  two 
equal  and  symmetrically  placed  loads.  The  bending-moment  dia- 
gram, Fig.  37,  and  the  shear  diagram,  Fig.  38,  are  left  as  studies 
for  the  student. 


1 

1 

tr 

jShear 

k 

\ 

tt<?  38 

7T 

39.  The  beam  of  Fig.  39  is  uniformly  loaded  with  w  pounds  per 
unit  of  length,  and  overhangs  the  supports  a  distance  a  at  each 
end.  Consider — quite  independently  of  each  other — the  two  over- 
hangs as  cantilevers  and  the  part  between  the  supports  as  a  simple 
beam. 

Since  the  overhangs  tend  to  bend  the  beam  concave  downward, 
the  bending  moments  due  to  them  are  negative.  The  bending  mo- 
ment of  each  overhang  cantilever  at  the  support  will  be  —  wa  X  s 

-c 

wa2 

-<rj- ,  and,  as  we  have  seen,  the  parabolic  curves  joining  the 

extremities  of  the  beam  and  the  ordinates  at  the  supports  represent 
the  bending-moment  diagrams  of  the  overhangs.     Disregarding  the 

load  on  the  central  span  of  the  beam,  the  bending  moment,  —  ^/- 

<> 

will  be  constant  between  the  supports.     Therefore,  the  negative  part 


THE  ELEMENTS  OF  THE  MECHANICS  OF  MATERIALS 


below  the  base  line  of  Fig.  40  represents  the  bending  moments  of 
the  whole  beam  due  to  the  overhangs. 


The  uniformly  loaded  central  span  of  length  I  occasions  a  reac- 
tion at  each  of  the  supports  of  *|- ,  and  as  the  tendency  is  to  bend 
the  beam  concave  upward  the  bending  moments  are  positive.  The 


BENDING  MOMENT. — SHEAR.  299 

bending  moment  at  any  section  between  the  supports,  distant  x 

-,f        wbx       wxz       wx  /7          N       ,  .  , 
irom  the  leit  support,  is  M  -     -~ —  —  -«—     :  -^-  (o — x)}  which, 

as  we  have  seen,  is  the  equation  of  a  parabola  having  its  axis  ver- 
tical under  the  middle  of  the  beam.  When  x  =  ~  we  have,  for  the 

maximum  bending  moment,  M max  =  -g-  .     The  parabola  above  the 

base  line  of  Fig.  40  is  the  bending-moment  diagram  of  the  part  of 
the  beam  between  the  supports,  and  is  positive.  By  superposition 
the  resultant  bending-moment  diagram  of  Fig.  41  is  obtained. 

The  shear  of  the  overhang  at  the  left  end  is  negative,  because  of 
its  contra-clockwise  tendency,  and  at  the  support  is  equal  to  —  wa. 
The  shear  of  the  overhang  at  the  right  end  is  positive,  because  of 
its  clockwise  sliding  tendency,  and  is  equal  to  wa  at  the  support. 
The  minus  and  plus  triangles  at  the  ends  of  Fig.  42  represent  the 
shears  of  the  left  and  right  overhangs,  respectively.  The  shear  at 

the  left  support  due  to  the  load  between  the  supports  is  -*-,  and  it 
decreases  an  amount  equal  to  w  for  each  unit  of  length  toward  the 
right  support.  At  the  middle  it  is,  therefore,^- —  w  ••*'=  0,  and 

at  the  right  support  it  is  -^ bw  — 5-.     The  shear  diagram 

of  Fig.  42  is  thus  obtained. 

EXAMPLE  I. — Figure  43  represents  a  beam  overhanging  both  sup- 
ports, and  carrying  a  uniformly  distributed  load  of  20  Ibs.  per  foot 
and  two  concentrated  loads  as  shown.  Find  the  support  reactions, 
the  maximum  positive  and  negative  bending  moments,  and  con- 
struct the  bending-moment  and  shear  diagrams. 

Solution:   Take  moments  about  the  supports  to  obtain  the  reac- 
tions ;  thus, 
lSRt  +  20  X  6  X  3  =  20  X  26  X  13  +  200  X  22  +  400  X  6. 

#±  =  7331  Ibs. 
18R2  -[-  20  X  8  X  4  +  200  X  4  =  20  X  24  X  12  +  400  X  12. 

R2  —  506|  Ibs. 

In  complicated  problems  like  this  it  is  the  best  practice  to  calculate 
the  bending  moments  at  various  sections  of  the  beam,  using  the 
results  as  ordinates  to  construct  the  diagram. 

For  the  bending-moment  diagram  let  the  linear  scale  be  0.1 
inch  =  1  foot,  and  the  load  scale,  1  inch  =  1000  Ibs.-ft. 


L    y 


BENDING  MOMENT.  —  SHEAR.  301 

The  bending  moments  of  the  overhangs  are  negative  because  of 
the  tendency  to  bend  the  beam  concave  downward. 
The  bending  moment  at  the  section  under  W^  is  : 

M  wi  =  -M>x4x2=-20X8=-160  Ibs.-f  t.  =  -  0.16  inch  to  scale. 
Jfjti  =-wX8x4-WiX4=-640-800=-  1440  Ibs.-ft.  =  -  1.44  inch. 
M3     =  R!  X  3  -  w  X  11  X  V  -  JPi  X  7  =  2200  -  1210  -  1400  =  -  4101bs.-ft.  =  -  0.41  inch. 
Jf4     =  JZj  x  4  -  10  X  12  X  6  -  Wi  X  8  =  2933^  -  1440  -  1600  =  -  106%  Ibs.-f  t.  =  -  0.107  inch. 
Jf5     =  ^  X  5  -  w  X  13  X  ¥  -  Wl  X  9  =  3666%  -  1690  -  1800  =  176%  Ibs.-ft.  =  0.177  inch. 
M9     =  Ri  X  9  -  w  X  17  X  ¥  -  Wl  X  13  =  6600  -  2890  -  2600  =  1110  Ibs.-f  t.  =  1.11  inch. 
Mwo  =  Si  X  12  -  w  X  20  X  10  -  Wl  X  16  =  8800  -  4000  -  3200  =  1600  Ibs.-f  t.  =  1.6  inch. 
-Ku    =  -Bi  X  15  -  w  X  23  X  ¥  -  Wi  X  19  -  TF2  X  3  =  710  Ibs.-f  t.  =  0.71  inch. 
MB*  =  RI  X  18  -  w  X  26  X  13  -  Wi  X  22  -  Ws  X  6  =  -  360  Ibs.-f  t.  =  -  0.36  inch. 

Using  these  scale  results  as  the  lengths  of  ordinates,  the  bending- 
moment  diagram  of  Fig.  44  was  constructed.  The  maximum  posi- 
tive bending  moment,  y,  is  under  W2  and  is  equal  to  1600  Ibs.-ft. 
The  maximum  negative  bending  moment,  y',  is  at  the  left  support, 
and  is  equal  to  1440  Ibs.-feet.  The  bending  moment  which  is  nu- 
merically the  greatest,  whether  positive  or  negative,  is  the  one  to  be 
considered  in  the  design  of  the  beam.  It  will  be  observed  that  the 
bending  moment  changes  sign  at  a  section  about  4^  feet  to  the  right 
of  the  left  support,  and  again  at  a  point  about  1  foot  to  the  left  of 
the  right  support.  It  will  be  shown  later  on,  when  treating  of  the 
deflection  of  beams,  how  some  bending  moments  between  the  sup- 
ports are  negative. 

For  the  shear  diagram  the  linear  scale,  0.1  inch  =  1  foot,  of  the 
bending-moment  diagram  will  be  used,  but  the  load  scale  will  be 
taken  as  1  inch  =  800  Ibs.,  so  that  1  sq.  inch  of  diagram  area  will 
equal  800  X  10  =  8000  Ibs.-ft. 

The  uniform  load  of  the  left  overhang  =  J^  =  —t  ^1|?  =    0.2 

oUU 


inch  to  scale. 

The  other  forces  affecting  the  shear  diagram  are  : 


_oqi  .     ,  .  TT2_400 

800~800~  ''  800~~800-  inch'  800  ~  800  ='~ 

_6X20  5i  nch.-gi-^i 

800  ~  "~800~  '800"~800~  800 


=  0.916  -  0.25  —  0.2  =  0.467  inch  ;  A  =  =  0.64  inch. 

oUU         oUU 

The  shear  of  the  left  overhang  is  negative  owing  to  its  contra- 

clockwise  tendency,  and  that  of  the  right  overhang  is  positive  from 

its  clockwise  tendency.     The  shear  of  the  left  overhang  is  —  W^ 

—  8w,  and  that  of  the  right  overhang  is  6w.     Construct  these  shears 

as  shown  in  Fig.  45.     The  shear  of  the  beam  between  the  supports 


OF  THE 

UNIVERSITY 


302        THE  ELEMENTS  OF  THE  MECHANICS  OF  MATERIALS 

is  R!  —  W2  —  ISw.  R!  is  made  up  of  two  component  parts — that 
due  to  the  left  overhang  and  that  due  to  the  load  between  the  sup- 
ports. The  resultant  positive  shear  at  the  left  support  is,  there- 
fore, R!  —  TF±  —  Sw  —  0.916  —  0.25  —  0.2  =  0.467  inch.  The 
shear  at  the  right  support,  due  to  the  loads  between  the  supports,  is 
(#!  —  Tf±  —  Sw)  --  W2  —  ISw  =  0.467  inch  —  0.5  inch  —  0.45 
inch.  Construct  the  component  parts  of  this  shear  as  shown  by 
the  shaded  portion  above  the  base  line  of  Fig.  45.  Revolve  the 
negative  part  of  the  shear  about  the  upper  boundary  of  the  posi- 
tive part  so  that  it  assumes  the  position  indicated  by  the  dotted 
lines  of  Fig.  45.  By  this  virtual  superposition  of  the  negative  and 
positive  parts  the  final  shear  diagram,  Fig.  46,  is  obtained. 

40.  Checks  as  to  the  accuracy  of  the  shear  diagram,  Fig.  46,  exist 
in  the  facts  that :  ( 1 )  The  sum  of  the  positive  areas  must  equal  the 
sum  of  the  negative  areas.  (2)  The  algebraic  sum  of  the  areas  to 
the  right  of  any  section  must  equal  the  algebraic  sum  of  the  areas 
to  the  left  of  the  section.  (3)  The  algebraic  sum  of  the  areas  to 
the  right  or  to  the  left  of  any  section  must  equal  the  bending 
moment  at  the  section  to  the  scale  of  the  diagram.  Thus  the  alge- 
braic sum  of  the  areas  of  the  shear  diagram,  Fig.  46,  to  the  right 
or  to  the  left  of  the  section  under  W2  is  found  by  measurement  to 
be  0.2  square  inch.  The  bending  moment  at  that  section  is,  there- 
fore, 8000  X  0.2  =  1600  lbs.-ft.,  as  was  found  above  by  calcula- 
tion. 

PROBLEMS. 

28.  A  uniform  beam  25  ft.  in  length,  whose  weight  is  disre- 
garded, is  supported  at  the  ends  and  has  a  concentrated  load  of 
400  Ibs.  at  9  ft.  from  the  left  support,  and  one  of  500  Ibs.  at  18  ft. 
from  the  left  support.     Construct  the  bending-moment  and  shear 
diagrams,  and  find  the  maximum  bending  moment  in  lbs.-ft. 

Ans.  3564  lbs.-ft. 

29.  Construct  the  bending-moment  and  shear  diagrams  of  the 
beam  of  Example  1,  the  weight  of  the  beam,  500  Ibs.,  being  con- 
sidered.    Find  also  the  maximum  bending  moment. 

Ans.  5112.5  lbs.-ft. 

30.  A  beam  12  ft.  long  is  supported  at  the  ends  and  loaded  with 
a  weight  of  3  tons  at  a  point  2  ft.  from  one  end.     Find  the  bending 
moment  at  the  middle  of  the  beam,  and  also  the  shearing  force. 

Ans.  3  tons-feet;   0.5  ton. 


BENDING  MOMENT. — SHEAR.  303 

31.  In  a  uniform  beam  of  length  L,  supported  at  the  ends,  and 
loaded  at  the  center  with  a  load  W,  show  that  the  bending  moment 

WL, 

is  greatest  at  the  middle  of  the  beam  and  equal  to  — ^-.     Then 

determine  graphically  the  bending  moment  and  shearing  force  at 
the  section  6  feet  from  either  support  of  a  beam  of  25  feet  span 
and  loaded  with  5  tons  at  the  middle. 

A  ns.  15  tons-ft. ;    ±  f  tons. 

32.  A  cantilever  projects  10  feet  from  a  wall  and  carries  a  uni- 
form load  of  60  Ibs.  per  foot;   it  also  supports  three  concentrated 
loads  of  100,  300,  and  500  pounds  at  distances  from  the  wall  of  2, 
5,  and  9  feet,  respectively.     Find  the  maximum  bending  moment, 
and  the   maximum   shear.     Construct  the   bending-moment   and 
shear  diagrams.  Ans.  —  9200  Ibs.  ft. ;  —  1500  Ibs. 

33.  A  beam  overhangs  both  supports  equally,  carries  a  uniform 
load  of  80  Ibs.  per  foot,  and  has  a  load  of  1000  Ibs.  in  the  middle, 
the  length  of  the  beam  being  15  feet,  and  the  distance  between  the 
supports  8  feet.     Construct  the  bending-moment  and  shear  dia- 
grams, and  find  the  maximum  bending  moment  and  the  maximum 
shear.  Ans.  2150  lbs.-ft;   820  Ibs. 


CHAPTER  III. 
MOMENT  OF  INERTIA.     RADIUS  OF  GYRATION. 

41.  Moment  of  Inertia. — The  moment  of  inertia  of  a  surface  is 
the  sum  of  the  products  of  each  elemental  area  of  the  surface  by 
the  square  of  its  distance  from  an  axis  about  which  the  surface  is 
supposed  to  be  revolving.     If,  instead  of  a  surface,  we  have  a  body, 
then  we  must  substitute  the  elemental  volume  for  the  elemental 
area  in  finding  the  moment  of  inertia.     The  moment  of  inertia 
varies  according  to  the  position  of  the  axis,  being  smallest  when 
the  axis  passes  through  the  center  of  gravity. 

It  would  be  a  crude  and  inaccurate  method  of  finding  the  moment 
of  inertia  by  actually  dividing  an  area  or  a  volume  into  its  ele- 
ments, multiplying  each  by  the  square  of  its  distance  from  the 
axis,  and  then  finally  taking  the  sum  of  these  products.  We  can, 
however,  find  the  moment  of  inertia  with  accuracy  by  means  of 
integration. 

The  strength  of  a  beam  or  of  a  column  depends  upon  the  form 
as  well  as  the  area  of  its  section,  and  in  all  calculations  respect- 
ing the  strength  of  beams  and  columns,  the  factor  which  gives  ex- 
pression to  the  effect  of  the  form  of  the  section  is  its  moment  of 
inertia  about  an  axis  passing  through  the  center  of  gravity. 

The  moment  of  inertia  of  a  surface  is  universally  denoted  by  I, 
the  axis  being  in  the  plane  of  the  surface  and  passing  through  its 
center  of  gravity.  When  the  axis  is  perpendicular  to  the  plane  of 
the  surface,  the  moment  of  inertia  is  then  termed  the  polar  moment 
of  inertia,  and  is  denoted  by  Ip. 

42.  There  are  certain  relations  between  the  different  moments  of 
inertia  of  the  same  section  which  are  useful  in  the  solution  of  prob- 
lems. 

43.  Let  I  denote  the  moment  of  inertia  of  any  surface  about  an 
axis  through  the  center  of  gravity  of  the  surface,  T  the  moment  of 
inertia  of  the  surface  about  any  other  parallel  axis,  A.  the  area  of 
the  surface,  and  H  the  perpendicular  distance  between  the  axes. 
Then  we  shall  have  P  =  I  +  AH2. 


MOMENT  OF  INERTIA.  —  EADIUS  OF  GYRATION 


305 


For,  let  yy,  Fig.  47,  be  the  axis  through  the  center  of  gravity, 
and  yy  the  axis  parallel  to  yy,  both  being  in  the  plane  of  the 
surface. 

Denote  the  elemental  areas  of  the  surface  by  a19  a2,  as,  etc.,  and 
their  distances  from  the  axis  yy  by  r±,  r2,  r3,  etc.  Then 

r  =  *'  —  #2     aH  —  r        aH     r        etc- 


etc. 


+  aB(H2  +  2rzH 
=  asl  +  a2rl+  asrl  +  H2  (a,  +  a,  +  a8) 


The  first  term  of  the  second  member  of  this  equation  is,  by  our 
definition,  the  moment  of  inertia,  I,  of  the  surface  about  the  axis 
yy  through  the  center  of  gravity.  The  second  term  becomes  AH2, 
in  which  A  denotes  the  whole  area.  The  third  term  will  reduce  to 
0,  because  the  quantity  within  the  parenthesis  is  the  algebraic  sum 
of  the  moments  of  the  elemental  areas  about  a  line  passing  through 
their  center  of  gravity.  Hence,  T  =  I  +  AH2. 

44.  Suppose  an  axis  perpendicular  to  the  plane  of  the  surface 
represented  in  Fig.  48  to  pass  through  0.  Let  a  be  an  elemental 
area  of  the  surface,  distant  r  from  0,  and  let  X  and  Y  be  rectangu- 
lar axes  passing  through  0  and  lying  in  the  plane  of  the  surface. 
Then  the  polar  moment  of  a  is  Ip  =  ar2.  The  moment  of  inertia 
of  a  about  X  is  Ix  =  ayz,  and  similarly  Iy  =  ax2.  But  we  have 
x2  +  f  —  r2,  therefore,  ax2  +  ay2  =  ar2.  That  is,  the  polar 
20 


306      THE  ELEMENTS  OF  THE  MECHANICS  OF  MATERIALS 

moment  of  inertia  of  any  surface  is  equal  to  the  sum  of  the  moments 
of  inertia  of  the  surface  about  any  two  rectangular  axes  lying  in 
the  plane  of  the  surface  and  passing  through  the  polar  axis  of 
revolution ;  that  is,  Ip  =  Ix-\-  Iy. 


45.  Radius   of   Gyration. — We  have  seen  that  1  =  ^rj-f-  azr\ 
+  a3rl  +  etc.,  in  which  ct1  +  &2  +  as  +  etc.  =  A,  the  whole  area. 
Now  if  we  can  conceive  the  whole  area  to  be  condensed  into  a  single 
particle,  distant  K  from  the  axis  of  rotation,  we  shall  have  I  =  AK2. 
This  imaginary  point  at  which  the  particle  is  supposed  to  be  situated 
is  known  as  the  center  of  gyration,  and  its  distance,  K,  from  the  axis 

is  the  radius  of  gyration.     We  have  then,  K  =  A/-T,  in  which  the 

mass,  M,  the  volume,  V,  or  the  weight,  W,  may  be  substituted  for 
the  area,  A. 

46.  If  we  denote  the  product  of  a  force,  area,  volume,  or  weight 
by  its  arm  as  the  first  moment,  or  simply  the  moment,  of  the  force, 
area,  volume,  or  weight,  then  we  may  conveniently  denote  the  pro- 
duct of  the  force,  area,  volume,  or  weight  by  the  square  of  its  arm 
as  the  second  moment  of  the  force,"  area,  volume,  or  weight.    Thus 
the  moment  of  inertia  is  sometimes  known  as  the  second  moment. 

EXAMPLE  I. — Find  the  least  moment  of  inertia  and  the  least  ra- 
dius of  gyration  of  a  parallelogram. 

The  least  moment  of  inertia  is  that  about  an  axis  through  the 
center  of  gravity,  g,  Fig.  49. 
The  elemental  area  =  2Bdh. 
Second  moment  of  the  elemental  area  =.  2B-hz-dh. 


MOMENT  OF  INERTIA.  —  EADIUS  OF  GYRATION  307 


Moment  of  Inertia 


/*~  7?  ff3 

ia  —  I  —  2B    I     tfdh  =  =^e-  - 
t/o  M 


IT-  H 

^- 


If  the  moment  of  inertia  with  respect  to  an  axis  coinciding* 
with  the  base  be  desired,  we  have  T  =  I  +  -41?  2,  in  which  JEf,  in 

rr 

this  instance,  =  -~  . 


And^T 


-/ 


H 
3 


EXAMPLE  II.  —  Find  the  polar  moment  of  inertia  and  radius  of 
gyration  of  a  circle  about  an  axis  passing  through  its  center. 
Elemental  area  =  %-rrrdr,  Fig.  50. 
Second  moment  of  elemental  area  =  2-jrr2rdr. 


R 


308      THE  ELEMENTS  OF  THE  MECHANICS  OF  MATERIALS 


EXAMPLE  III.  —  Find  the  polar  moment  of  inertia  and  radius  of 
gyration  of  a  cone  about  its  axis. 
Elemental  volume  =  7rr2dh,  Fig.  51. 

r" 
X 


Second  moment  of  elemental  volume  =  VK2  = 
for  the  circle,  K2  =  ^  . 

7T       i*R 

Then  /  =  r»   I  r*dh.     By  similar  triangles,  we  have 
^t/o 

h       r       ,  7        ^Tr        j  77.       ffdr 

~ff=  -n>  whence  h  =  -„  -  ,  and  aA  =  —  „-  . 


PEOBLEMS. 

Find  the  moment  of  inertia  and  radius  of  gyration  of  the  follow- 
ing: 

34    Triangle  about  an  axis  through  vertex  and  parallel  to  base. 


T,  rr          H        - 

Ans.     1    =  —  3  —  ;  XL  =  -jr-  ^/^  . 

4  /i 


35.  Triangle  about  an  axis  through  the  center  of  gravity  and 

parallel  to  base. 

BH*.  K=H    .VTT 
36    '          "  18 


Ans.     /  = 


MOMENT  OF  INERTIA. — KADIUS  OF  GYRATION  309 


36.  Triangle  about  an  axis  coinciding  with  base. 

A  *• -£,>:•; 

37.  Trapezoid  about  an  axis  coinciding  with  small  base. 

Ans.     I'  =  ^ ;  K  —  j 


38.  Trapezoid  about  an  axis  coinciding  with  large  base. 

A  Tf  •*-*•     \&U  ~T~   *-*)          1? 

JLns.     A    =•  .,  - 5  J$.  == 


39.  Trapezoid  about  an  axis  through  center  of  gravity  and  par- 
allel to  base. 


40.  Square  about  its  diagonal. 

Am.    1  =  j3>K  =  & 

41.  Circle  about  a  diameter. 

Ans.    1='^—--  K=—-. 

42.  Hollow  circle  about  a  diameter. 


.  /=*(!*-#);  K  = 


+ 


64 v  4 

Find  the  polar  moment  of  inertia  and  radius  of  gyration  of  the 
following  surfaces : 

43.  Parallelogram  about  a  pole  passing  through  its  center  of 
gravity. 

44.  Circle  about  a  pole  passing  through  its  center. 

Ans.     Ip  =  --g- ;  K  —  •—  </$  . 

45.  Hollow  circle  about  a  pole  passing  through  center. 

Ans.    Ip  =  ~  (Z>4  -d4)-K=  J^^-  • 

Find  the  polar  moment  of  inertia  and  radius  of  gyration  of  the 
following  named  solids : 

46.  Cylinder  about  a  pole  coinciding  with  its  axis. 

Ans.     ^  =  ^rx^=~V8. 


310      THE  ELEMENTS  OF  THE  MECHANICS  OF  MATERIALS 

47.  Hollow  cylinder  about  a  pole  coinciding  with  its  axis. 

Ans.     lr  =  ^(iy-^;K 

48.  Sphere  about  a  diameter. 

An,    4. 

49.  A  bar  of  rectangular  section  about  a  pole  passing  through  the 
center  of  figure. 


CHAPTER  IV. 
THE  THEORY  OF  BEAMS. 

47.  Strain.  —  The  change  of  form  which  a  load  produces  in  a 
body  is  called  the  strain  due  to  the  load. 

48.  Stress.  —  The  internal,   or  molecular,   resistance  which  the 
material  of  a  body  interposes  to  resist  deformation  is  called  a  stress. 

49.  Coefficient   of  Elasticity.  —  Within  the  elastic  limit  of  all 
materials  the  stresses  are  proportional  to  the  strains,  but  since  the 
same  intensity  of  stress  does  not  produce  the  same  strain  in  different 
materials  we  must  have  some  definite  means  of  expressing  the 
amount  of  strain  produced  in  a  body  by  a  given  stress.     The  means 
employed  is  to  assume  the  body  to  be  perfectly  elastic  and  then 
state  the  intensity  of  stress  necessary  to  strain  the  body  by  an 
amount  equal  to  its  own  length.     This  stress  is  known  as  the 
Modulus   of   Elasticity,   or  the   Coefficient  of   Elasticity,   and   is 
denoted  by  E.     The  values  of  E  for  different  materials  have  been 
determined  and  are  practically  the  same  for  tension  and  compres- 
sion.    The  mean  values  for  E  in  pounds  per  square  inch  for  the 
materials   most   commonly   used   in   engineering   are   as   follows: 
Timber,  1,500,000;  cast  iron,  15,000,000  ;  wrought  iron,  25,000,000; 
steel,  30,000,000. 

There  are  no  materials  of  construction  that  are  perfectly  elastic, 
and  but  few  that  will  stretch  0.001  of  their  length  and  remain 
elastic. 

50.  If  y  denotes  the  strain  produced  in  a  body  by  a  stress  S, 
and  Y  the  strain  produced  by  the  stress  necessary  to  stretch  the 
body  to  double  its  length;  then,  since  the  stresses  are  proportional 
to  the  strains,  we  shall  have  for  a  stress-strain  diagram,  Fig.  52, 


=     ,  whence  E  =        = 

E'  y       y 

since  the  strain,  or  deformation,  F,  is  equal  to  L. 


312      THE  ELEMENTS  OF  THE  MECHANICS  OF  MATERIALS 


51.  Neutral  Axis. — When  a  beam  is  deflected,  as  in  Fig.  53,  its 
fibers  on  the  convex  side  are  in  tension  and  those  on  the  concave 
side  in  compression.  There  must  be  a  surface  somewhere  between 
these  conditions  where  the  fibers  are  neither  in  tension  nor  in  com- 


pression.  Such  surface  is  known  as  the  neutral  surface,  and  the 
line  in  which  this  surface  intersects  the  plane  of  a  section  of  the 
beam  is  the  neutral  axis  of  the  section.  The  neutral  axis  invariably 
passes  through  the  center  of  gravity  of  the  section. 


rig.  S3 

52.  Suppose  an  elastic  beam  to  be  bent  to  the  arc  of  a  circle, 
Fig.  54.  Let  r  denote  the  radius  of  the  neutral  surface,  and  c  the 
distance  from  the  neutral  surface  to  the  outermost  surface. 


The  length  of  the  neutral  surface  will  not  change  in  the  bending, 
hence  %TTT  =.  original  length  of  the  outermost  surface,  and  %ir(r-\-c) 
=  length  of  outermost  surface  after  bending. 


THE  THEORY  OF  BEAMS 


313 


The   deformation  =  strain   of   outer   surface  =  2ir(r  -\-  c) — 2irr 
=  27rc.     But  we  have  seen  that 

Strain  _Stress_  _  _S 

E' 


Original  Length 


Hence 


Modulus  of  Elasticity 

£__-C  __  S_ 

r~r  ~  E  ' 


(A) 


-pi  o 

Then,  from  equation  (A),  we  have  —  =  -  —  unit  stress  at  a 

T  C 

distance  unity  from  the  neutral  surface,  8  being  the  unit  stress, 
tension  or  compression,  of  the  fibers  of  the  outermost  surface,  and  c 
the  distance  from  the  neutral  surface  to  the  outermost  surface. 


.  SS. 


53.  Resisting  Moment.  —  Suppose  a  beam  section,  Fig.  55,  to  be 
made  up  of  a  great  number  of  layers  of  areas  a,  aly  a2,  &c.,  distant 
y>  y\>  2/2?  &c.,  from  the  neutral  axis.  The  unit  stresses  at  distances 

o  o  cr 

y>  Vi>  y^  etc.,  are  -  X  y,    -  X  y,,   -  X  ya,  and  the  total  stresses  on 


o  rr  nr 

the  elemental  areas  are  -  X  ay,    -  X  a^,   -  \  X 

CO  C 

oo 

moments  of  these  stresses  are  -  X   ay\  -  X  c^y 

c  c 


^z,  &c.      The 

o 

-  X  azyl,  &c. 

c 

The  sum  of  all  these  moments  will  be  the  resisting  moment  of  the 

o 

section,    hence    Kesisting   Moment  =   -  (ay2  +  0^1  +  a$\  +  &c.) 

c  * 


RfJ" 


=  -—  ,  in  which  I  is  the  moment  of  inertia  of  the  section  about 

C/ 

the  neutral  axis.  The  Bending  Moment  must  be  equal  to  the 
Resisting  Moment,  since,  for  equilibrium,  the  internal  stresses 
must  equal  the  external  forces,  therefore, 


Bending  Moment  =  M  = 


(B) 


314      THE  ELEMENTS  OF  THE  MECHANICS  OF  MATERIALS 

54.  Section  Modulus.  —  The  factor  -  contains  the  dimensions  of 

c 

the  section  and  is  the  measure  of  its  strength.    It  is  called  the  modu- 
lus of  the  section,  and  is  denoted  by  Z.     Hence,  I  =  Zc,  and 

-o-  =  -  =  Z,  whence  S  =  -^  .     Substituting  this  value  of  8  in 
equation  (A),  we  have,  -  =  -^.,  whence  M  =  -  ==  —  .          (0) 


The  equation  (C)  just  found  is  that  of  the  elastic  curve,  showing 
the  relation  between  the  bending  moment  at  any  section  and  the 
radius  of  curvature  of  the  beam. 

55.  In  the  theory  of  beams  the  important  assumptions  made  are  : 
(a)   That  the  strain  increases  directly  as  the  distance  from  the 
neutral  axis.     (&)   That  the  stress  is  proportional  to  the  strain. 
(c)  That  the  coefficient  of  elasticity  is  the  same  for  tension  as  for 
compression. 

Experiments  have  shown  that,  within  the  elastic  limit  of  any 
material,  the  above  assumptions  may  be  regarded  as  perfectly  true. 

56.  The  equation,   M  =  -  -   is   the   fundamental   equation   for 

c 

beam  investigations,  and  since  I  is  expressed  in  bi-quadratic  inches, 
M  must  be  expressed  in  pounds-inches.  The  length  of  a  required 
beam,  and  the  load  to  which  it  is  to  be  subjected  being  known,  we 
can  readily  find  the  maximum  bending  moment,  and  then  by  assum- 
ing the  allowable  working  stress,  8,  per  unit  of  area  at  the  outer- 

most fiber,  we  shall  have,  -«-  =  -.     The  numerical  value  thus  ob- 

tained for  -  must  be  satisfied  by  the  dimensions  assumed  for  the 

c 
cross-section  of  the  beam.     The  smaller  value  of  8,  whether  for 

tension  or  compression,  should  be  taken. 

57.  Dangerous  Section.  —  The  section  of  a  beam  at  which  the 
bending  moment  is  a  maximum  is  known  as  the  "  dangerous  sec- 
tion." 

In  the  case  of  a  beam  uniformly  loaded  it  has  been  shown  that 
the  bending-moment  curve  is  parabolic  and  that,  therefore,  the 
ordinate  representing  the  bending  moment  is  a  continuous  func- 
tion of  x.  The  determination  of  the  dangerous  section  would  then 
be  to  find  that  value  of  x  which  would  make  the  bending  moment 
a  maximum.  To  do  this  we  would  place  the  first  ^-derivative  of  M 


THE  THEORY  OF  BEAMS  315 

equal  to  zero,  and  the  resulting  value  of  x  would  be  the  abscissa  of 
the  dangerous  section. 

For  example,  the  beam  of  Fig.  56  is  uniformly  loaded  with  w 

pounds  per  linear  foot.     The  support  reactions  are  each  -^  . 

The  bending  moment  at  any  section  distant  x  from  the  left  sup- 
port is, 

wLx       WX" 


Then,  -^  =  ?^  --  wx  —  0,   whence  %=-%• 
That  this  value  of  x  makes  M  a  maximum  is  evidenced  by  the 
negative  result  of  the  second  ^-derivative,  thus  : 


=  —  w. 


Fif.  56. 

The  maximum  bending  moment,  and  therefore  the  dangerous 
section,  is  then  at  the  middle  of  the  beam;  see  Fig.  27,  p.  291. 
It  will  be  observed  that  the  second  member  of  the  equation. 

—j-  =  -s wx,  is  the  shear  at  the  section  distant  x  from  the  left 

support.  We  infer  from  this  that  the  first  ^-derivative  of  the 
moment  is  the  shear,  and  that,  in  the  case  of  uniform  loading,  the 
maximum  bending  moment  is  at  the  section  where  the  shear  is 
zero;  see  Fig.  29,  p.  291. 

If,  in  addition  to  the  uniformly  distributed  load,  the  beam  of 
Fig.  56  were  subjected  to  one  or  more  concentrated  loads,  or  if 
there  were  no  uniform  load  and  the  beam  subjected  simply  to  one 
or  more  concentrated  loads,  the  ordinate  representing  the  bending 
moment  would  no  longer  be  a  continuous  function  of  x.  The  first 
derivative  of  the  moment  at  any  section  would,  however,  still  be 
the  shear  at  that  section,  but  the  shear  at  the  dangerous  section 
might  nofc  be  zero  because  the  shear  is  not  necessarily  zero  at  any 
section.  If  the  construction  of  the  shear  diagram  shows  that  the 
shear  is  not  zero  at  any  section,  it  will  also  show  that  at  some  one 


316      THE  ELEMENTS  OF  THE  MECHANICS  OF  MATERIALS 

section  the  shear  suddenly  changes  sign,  or  passes  through  zero, 
and  the  maximum  bending  moment  will  be  found  at  that  section; 
see  Fig.  24,  p.  289. 

The  construction  of  the  shear  diagram  will  at  once  locate  the 
dangerous  section,  or  it  may  be  found  by  taking  the  algebraic  sum 
of  the  forces  to  the  left  of  various  sections  until  the  point  is  found 
where  the  shear  changes  sign. 

EXAMPLE  I. — A  yellow  pine  beam  of  18  feet  span  is  to  support 
concentrated  loads  of  1200  Ibs.  and  1500  Ibs.  at  points  distant  4 
feet  and  10  feet  from  the  left  support.  Determine  the  cross-sec- 
tional dimensions  of  the  beam. 

From  Fig.  57  we  have:  18^  =  1200  X  14  +  1500  X  8,  whence, 
£±  =  1600  Ibs.  Also,  18R2  =  1500  X  10  +  1200  X  4,  whence, 
R2  =  1100.  The  construction  of  the  shear  diagram  shows  the  dan- 
gerous section  to  be  under  the  load  of  1500  Ibs.  Hence, 


10' 

1500 


Mmax  —  1600  X  10  —  1200  X  6  =  8800  Ibs.-ft.  =  105,600  Ibs.- 
inches.     Using  a  factor  of  safety  of  10,  we  shall  have  : 

9000 
S  =  -JQ-  —  900  Ibs.  per  square  inch.     Then, 

/__  105,600  _117oo 

-  ~~-    L17-33- 


Assuming  a  5"  x  12"  beam,  we  shall  have  : 

5  X  1728 


_  12ft 


c  ~    12c  ~     12  X  6 

This  is  so  near  the  assumed  value  of  the  section  modulus  that  it 
would  appear  that  the  5"  x  12"  beam  would  be  secure.  We  shall, 
however,  investigate  the  influence  of  the  weight  of  the  beam  itself. 
Weight  of  beam  =  5  X  12  X  if  X  -1/  =  300  Ibs.  =  16|  Ibs.  per 
foot,  and  #±  then  becomes  1600  +  150  =  1750  Ibs.  The  consid- 
eration of  the  weight  of  the  beam  will  not  change  the  position  of 
the  dangerous  section,  as  can  be  shown  on  the  shear  diagram.  Then 


THE  THEORY  OF  BEAMS  317 

Mmax  =  1750  X  10  —  1200  X  6  —  16.66  X  10  X  5  =  9466.66 
Ibs.-ft.  =  113,600  Ibs.-inches. 

Mo       113.600  X  6  X  12  .     ,       m,  . 

Then,  8  =  -j-  —  -      5  x  1728  =  Per  sq*  m 

is  greater  than  the  assumed  safe  stress  of  900  Ibs.,  so  a  5.5"  x  12" 

beam  will  be  selected.     Then, 

Weight  of  beam  =  5.5  X  12  X  H  X  -^  =  330  Ibs.  =  18.33  Ibs. 

per  foot,  and  R±  =  1765  Ibs. 

Then,  Mmax  =  1765  X   10  -  -  1200  X   6  -  -  18.33   X   10  X   5 

=  9534.33  Ibs.-ft.  =  114,412  Ibs.-inches. 

Then,    fl==U»X 


result  which  shows  the  5.5"  x  12"  beam  to  be  safe. 

58.  Standard  I  Beams.—  The  standard  steel  I  beams  so  exten- 
sively used  in  engineering  work  are  rolled  in  light,  intermediate, 

.  and  heavy  weights  of  thirteen  different  sizes.  The  different  manu- 
facturers issue  hand-books  containing  tables  of  the  properties  of  the. 
beams  they  produce,  and  while  there  is  an  agreement  in  sizes  as  to 
the  depth,  there  are  marked  differences  in  the  proportions  of  cross- 
sections,  and  therefore  also  of  the  weights  per  foot  and  moments  of 
inertia. 

59.  The  ultimate  strength  of  structural  steel  may  be  taken  as 
65,000  Ibs.  per  square  inch,  and  the  elastic  limit  at  about  one-half 
the  ultimate  strength.     The  working  fiber  stress  per  sq.  inch  may 
be  taken  as  16,000  Ibs.  for  buildings  and  12,500  Ibs.  for  bridges, 
indicating  factors  of  safety  of  4  and  5.2,  respectively. 

60.  The  span  and  load  of  an  I  beam  being  given,  the  value  of  i 

0 

may  be  found  by  means  of  the  formula,    —  =  -~-,  and  then  the  beam 
in  the  manufacturer's  tables  which  has  an  -     equal  to  or  next 

0 

greater  than  the  one  found  is  to  be  selected. 

EXAMPLE  II.  —  What  Cambria  I  beam  should  be  selected  for  a 
floor  bearing  a  load  of  180  Ibs.  per  square  foot,  the  beams  to  have 
a  span  of  25  feet,  spaced  8  feet  apart,  and  to  have  a  maximum 
unit  stress  of  16,000  Ibs.  per  sq.  inch? 

25  X  8  X  180  =  36,000  Ibs.  load  on  each  beam  =  ^%^°  =  1440 

/*o 

Ibs.  per  foot.     R^  —  18,000  Ibs.,  and 


318      THE  ELEMENTS  OF  THE  MECHANICS  OF  MATERIALS 

Mmax  =  18,000  X   12-5  X   12  -  -  1440  X   12.5   X  6.25  X 
=  1,350,000  Ibs.-in. 


A  reference  to  the  hand-book  of  the  Cambria  Steel  Co.  shows 
that  their  15-inch  special  I  beam  of  65  Ibs.  per  foot  should  be 
selected. 

EXAMPLE  III. — What  is  the  proper  size  of  I  beam  to  carry  a  load 
of  35,000  Ibs.  concentrated  at  the  middle  of  a  span  of  25  feet,  the 
fiber  stress  not  to  exceed  16,000  Ibs.  per  square  inch? 
Mmax  =  17,500  X  12.5  X  12  =  2,625,000  Ibs.-inches. 

L  —  K—  2,625,000  _  161  56 
c  ~  IS  "      16,000 

This  value  of  the  section  modulus  would  indicate  that  a  24-inch 
standard  Cambria  I  of  80  Ibs.  per  foot  would  do.  We  shall,  how- 
'ever,  investigate  the  effect  of  the  weight  of  the  beam  itself. 
Weight  of  beam  =  80  X  25  =  2000  Ibs.,  therefore  £±  =  18,500  Ibs. 
Then  Mmax  =  18,500  X  12.5  X  12  —  80  X  12.5  X  6.25  X  12 

=  2,700,000  Ibs.-inches.  8  =  ^  =  2?7°°^g^  12-  —  15,500, 
which  is  500  Ibs.  per  sq.  inch  less  than  the  safe  allowable  stress. 
(The  moment  of  inertia  of  the  light  24-inch  Cambria  standard  I  is 
2087.2.)  The  24-inch  standard  I,  weighing  80  Ibs.  per  foot,  is, 
therefore,  the  proper  selection. 

EXAMPLE  IV.  Heavy  12-inch  Cambria  I  beams  of  20  feet  span 
are  to  be  used  to  support  a  floor  bearing  a  uniform  load,  including 
the  weight  of  the  beam,  of  200  Ibs.  per  square  foot.  The  fiber  stress 
is  not  to  exceed  16,000  Ibs.  per  sq.  inch.  Find  the  spacing  of  the 
beams. 

From  the  table  of  properties  of  standard  Cambria  I  Yearns,  we 
find  for  the  beam  selected, 

Section  Modulus  =  -  =  41. 
c 

Let  x  denote  the  distance  between  center  lines  of  consecutive  beams. 
Then  2Qx  X  200  =  4000:r  =  load  on  one  beam,  and  — <^p  =  200z 
=  load  per  foot.     R^  —  2000z. 
Mmax  —  2000z  X  10  X  12  —  200z  X  10  X  5  X  12  =  120,000z. 

Then  i  =  f,  or  41  =  -?!!'nn!r  >  whence  x  =  5-47  feet- 
C        o  lo,UUU 


THE  THEORY  OF  BEAMS  319 

The  load  on  each  beam  is  then  4000  X  5.47  =  21,880  Ibs.  But 
this  includes  the  weight  of  the  beam,  which  is  40  X  20  =  800  Ibs. ; 
hence,  21,080  Ibs.  may  be  placed  on  the  beam. 

PROBLEMS. 

50.  A  rectangular  beam,  9  inches  deep,  3  inches  wide,  supports  a 
load  of  0.5  ton,  concentrated  at  the  middle  of  an  8-foot  span.     Find 
the  maximum  fiber  stress.  Ans.  0.3  ton  per  sq.  inch. 

51.  A   beam   arranged   with   symmetrical   overhanging   ends   is 
loaded  with  three  equal  loads — one  at  each  end  and  one  at  the  mid- 
dle.    What  is  the  distance  apart  of  the  supports,  in  terms  of  the 
total  length,  L,  when  the  bending  moment  is  equal  over  the  supports 
and  in  the  middle,  and  at  what  section  is  the  bending  moment  zero  ? 

Ans.  -g-.    At  sections  between  the  supports  distant  -^  from  them. 

52.  A  square  timber  beam  of  12-inch  side  weighs  50  Ibs.  per  cubic 
foot,  is  20  ft.  long,  and  supports  a  load  of  2  long  tons  at  the  middle 
of  its  span.     Calculate  the  fiber  stress  at  the  middle  section. 

Ans.  1037  Ibs.  per  sq.  inch. 

53.  Light  12-inch  Cambria  I  beams  of  20  feet  span  are  to  sup- 
port a  floor  bearing  a  uniform  load  of  175  Ibs.  per  square  foot.   The 
outermost  fiber  stress  is  not  to  exceed  16,000  Ibs.  per  sq.  inch.     At 
what  distance  apart  should  the  beams  be  placed  ? 

Ans.  5.5  ft. 

54.  How  much  stronger  is  a  beam  4  inches  wide,  8  inches  deep, 
and  8  ft.  long,  than  one  3  inches  wide,  5  inches  deep,  and  14T^-  ft. 
long?  Ans.  Six  times  as  strong. 

55.  What  safe  uniform  load  can  be  placed  on  a  wooden  cantilever 
5  feet  long,  3  inches  wide,  and  5  inches  deep,  in  order  that  the  fiber 
stress  shall  be  900  Ibs.  per  square  inch? 

Ans.  71  Ibs.  per  foot,  nearly. 

56'.  What  safe  uniformly  distributed  load  can  be  placed  on  a 
standard  Cambria  channel  beam  weighing  20.5  Ibs.  per  foot,  the 
span  being  16  feet,  the  web  placed  vertical,  and  the  maximum  fiber 
stress  not  to  exceed  12,500  pounds  per  square  inch  ? 

Ans.  11,120  Ibs. 

57.  Find  the  factor  of  safety  of  a  heavy  12-inch  Cambria  I  beam 
of  15  feet  span  when  sustaining  a  uniformly  distributed  load  of 
30  tons.  Ans.  2. 


CHAPTER  V. 
COLUMNS.     SHAFTS. 

61.  Columns. — A  column,  or  strut,  is  a  straight  beam  acted  on 
compressively  at  its  extremities,  and  of  such  length  compared  with 
its  diameter  or  least  sectional  dimensions,  that  failure  will  result 
from  buckling,  or  lateral  bending,  instead  of  by  crushing  or  by 
splitting.     In  addition  to  the  many  familiar  applications  of  col- 
umns in  structural  work,  the  piston-rods  and  connecting-rods  of 
steam  engines  are  also  classed  as  columns. 

62.  If  columns  were  initially  absolutely  straight,  made  of  homo- 
geneous material,  and  exactly  centrally  loaded,  there  would  be  no 
difference  in  the  character  of  their  failure  from  that  of  short  speci- 
mens.    These  three  conditions  are  never  fulfilled  in  practice,  and 
in  consequence  columns  are  weaker  than  short  blocks  of  the  same 
material.     No  satisfactory  theoretical  discussion  of  columns  has 
been  made,  and  all  the  formula}  used  in  their  design  contain  em- 
pirical constants  determined  by  experiment.     The  formula  having 
the  most  rational  basis  is  the  one  attributed  to  Rankine  or  to  Gor- 
don, and  is, 

AS 

(D) 

aK* 

in  which  P  is  the  load  on  the  column  expressed  in  pounds,  A  the 
sectional  area  of  the  column  in  square  inches,  8  the  crushing 
strength  of  a  short  block  of  the  material,  a  a  constant  quantity 
determined  by  experiment  for  different  materials,  L  the  length  of 
the  column  in  inches,  and  K  the  radius  of  gyration  of  the  section. 

63.  The  strength  of  a  column  depends  largely  upon  the  manner 
in  which  its  ends  are  secured.     If  the  ends  are  square,  or  flat,  the 
column  is  said  to  have  fixed  ends ;  if  one  end  be  fixed  and  the  other 
end  hinged,  as  in  the  case  of  a  piston-rod,  the  end  conditions  are 
known  as  "  pin  and  square ;  "  if  both  ends  be  hinged  or  "  rounded," 
as  in.  the  case  of  a  connecting-rod,  the  end  conditions  are  known  as 
"pin,"  or  "round."     The  value  of  K2  in  formula  (D)   is  found 


COLUMNS. — SHAFTS 


321 


from  the  relation  7  =  AK2,  and  the  values  for  a,  for  8,  and  for 
suitable  factors  of  safety  for  the  three  conditions  of  bearing,  are 
given  in  the  following  table.  It  will  be  observed  from  the  table 
that  the  ratios  of  the  values  of  a  for  the  fixed  ends  to  the  values  of 
pin  and  square  and  pin  ends,  are  as  1.78  to  1  and  4  to  1,  respec- 
tively, and  these  ratios  indicate  the  relative  strengths  of  long  col- 
umns with  these  end  conditions. 


Timber. 

Cast  Iron. 

Wrought 
Iron. 

Structural 
Steel. 

Hard 
Steel. 

s 

a  (Fixed  Ends)  .... 

8000 
3000 

84000 
5000 

55000 
6000 

50000 
36000 

150000 
25000 

a  (Pin  and  Square)  
a  (Pin)  

1690 
750 

2810 
1250 

20250 
9000 

24000 
18000 

14060 
6250 

Factor  of  Safety  for 
Buildings  

8 

8 

6 

4 

5 

Factor  of  Safety  for 
Bridges  

10 

10 

5 

7 

Factor  of  Safety  for 
Shocks  

15 

15 

15 

EXAMPLE  I.  —  A  hollow  cylindrical  cast  iron  column,  16  ft.  long, 
with  fixed  ends,  sustains  a  load  of  200,000  Ibs.  when  used  in  a 
building.  Outside  diameter,  10  inches;  inside  diameter,  8  inches. 
Is  it  safe  ? 

A  =  0.7854(100  —  64)  =  28.27  sq.  inches. 
-  4096)  _  1ft  9, 


64  x  28.27 


P  = 


28.27 


incn' 


The  factor  of  safety  is   g^-    =10  nearly,  which  shows  the  column 
to  be  perfectly  safe. 

EXAMPLE  II.  —  Find  the  safe  load  for  a  light  12-inch  Cambria 
I  column,  weighing  31.5  lbs.  per  foot,  16  feet  long,  and  with  fixed 
ends,  the  column  to  be  used  in  a  building. 
21 


322      THE  ELEMENTS  OF  THE  MECHANICS  OF  MATERIALS 

Eeferring  to  the  Cambria  hand-book,  we  find  that  A  =  9.26,  and 
K  =•  1.01.  The  factor  of  safety  is  4.  Hence, 

p_       SA  12,500  x  9.26      _  12,500  x  9.26  x  36,360  _  5?  ?55 

1         Z2   ~  j  (16  x  12)2  36,360  +  367864 

aK'1  36,000  x  1.02 

per  square  inch. 

64.  In  designing  a  column  we  have  given  the  form  of  its  section, 
its  length,  the  material  of  which  it  is  made,  the  load  it  is  to  carry, 
and  the  manner  of  securing  its  ends.  The  problem  is  then  to  deter- 
mine the  necessary  area  of  cross-section.  The  general  method  of 
procedure  is  to  determine  from  the  data  given  the  necessary  cross- 
section  area  for  a  short  block  of  the  material;  then,  knowing  that 
the  section  area  of  the  required  column  must  be  larger,  assume 
dimensions  which  give  a  larger  area,  and  then,  by  means  of  formula 
(D),  ascertain  the  unit-stress  resulting  from  such  assumption.  If 
it  is  too  great,  the  assumed  area  is  too  small ;  if  too  little,  a  smaller 
area  must  be  chosen.  Proceed  thus  by  trial  and  error  until  a  satis- 
factory solution  is  obtained. 

EXAMPLE  III. — A  column  of  timber  of  square  section,  14  ft. 
long  and  with  fixed  ends,  is  required  to  support  a  steady  load  of  10 
tons.  Find  the  dimensions  of  the  section. 

8000 
Using  a  factor  of  safety  of  8,  we  have  — g—  :=  1000  Ibs.  per  sq. 

inch  as  the  safe  working  unit  stress.    Hence,  — ^™ —  —  20  square 


inches  area  of  section  needed  for  a  very  short  column,  or  one  less  in 
length  than  ten  times  the  least  dimension  of  the  section.  As  the 
required  section  must  be  larger,  we  will  assume  an  area  of  36  square 

inches.     Then  A  =  (6)2,  and  K2  =  ~  =  l2~x^6V  =  3'     Hence> 


per  square  inch.    Since  this  is  much  larger  than  the  allowable  unit 
stress  of  1000  Ibs.,  we  must  assume  a  larger  area.     Try  an  area  of 

64  square  inches.     Then  A  =  (8)  2,  and  I  =  12^(SY  =  ^  • 

T-T          a      20,000  /  ..    ,   (15  X  12)2  X  3  1 

Hence,  S=     '64    1  1  +     3000  x  16     J  ~  Per  S(luare  m- 

This  is  a  trifle  small,  so  we  will  try  a  square  area  whose  side  is  7.9 
inches.     Then  A  =  (7.9)2,  and  K2  =  12          .v    =  6'2  ' 


COLUMNS.  —  SHAFTS  323 

0       20,000  f  1     ,    (15  X  12)2\ 
Hence,  8  =-  -TJfjfy  j  l  ~r  SQQQ  x  5  2  J  ~  ~  Per  S(luare  mcn> 

which  is  quite  near  the  allowable  unit  stress.     Therefore,  a  column 
of  square  section,  haying  a  side  of  7.9  inches,  will  suffice. 

65.  Sometimes  all  the  dimensions  can  be  assumed  except  one,  and 
then,  after  expressing  A  and  K2  in  terms  of  the  unknown  dimen- 
sion, we  can  substitute  them  in  formula  (D)  and  solve  the  resulting 
bi-quadratic  equation  for  the  unknown  dimension.  Thus,  in  Ex- 
ample III,  let  x  denote  the  side  of  the  unknown  square  section. 

x4         x2 
Then  A  —  x2,  and  K2  =  ^-2  =  ^  .     Substituting  these  values  in 

(D),  we  shall  have: 

i  poo  -  20,000  J1  j_  (15  X12)2X  12  \_  20,000  /,    ,   129.6\ 

~~^l  3000  x  ^     /  =  ~^~l      Tr 

from  which  we  get,  x  =  7.8  inches. 

EXAMPLE  IV.  —  A  hollow  cylindrical  cast  iron  column,  20  feet  in 
length,  is  required  to  sustain  a  steady  load  of  164,000  pounds.  De- 
termine its  cross-sectional  dimensions. 


Using  a  factor  of  safety  of  7,  we  have  —  12,000  Ibs.  per 

square  inch  as  the  safe  unit  working  stress.  Then  1  '  =  13.66 
square  inches  of  area  needed  for  a  very  short  column.  Assuming  an 
area  of  25  square  inches,  and  assuming  further  that  the  outside 
diameter  of  the  column  shall  be  10  inches,  we  shall  have,  calling 
d  the  inside  diameter,  0.7854  X  100  —  0.7854  X  d2  —  25,  whence 


d  =  8.126  inches.     Then  K2  =        =  ^ 

-A. 


}  =  13,8,0  Ib,  per 


Her.ce  *= 

inch,  which  is  greater  than  the  allowable  unit  stress  of  12,000  Ibs. 

Assuming  an  area  of  29  square  inches,  we  shall  have  : 
0.7854  X  100  —  0.7854  X  d2  =  29,  whence  d  =  7.94  inches,  and 
then  K2  =  10.19. 

Hence  S  =  ^^{  !  +  JML  }  12)048  Ibs.  per  „.  inch, 
a  result  sufficiently  near  the  allowable  stress  to  warrant  making 
the  column  10  inches  in  outside  diameter  and  the  metal  1  inch 
thick. 

66.  For  the  properties  of  the  different  forms  of  the  built-up 
beams  and  columns  so  largely  used  in  structural  work,  reference 


324      THE  ELEMENTS  OF  THE  MECHANICS  OF  MATERIALS 

should  be  made  to  the  hand-books  issued  by  the  different  steel 
companies. 

67.  Shafts.  —  When  cylindrical  shafts  are  used  for  the  transmis- 
sion of  power  they  are  subjected  to  torsion,  or  twist,  which  occa- 
sions a  shearing  stress.     The  angle  through  which  the  elements  are 
twisted  is,  within  the  elastic  limit  of  the  material,  proportional  to 
the  applied  force.     The  elements  along  the  axis  of  the  shaft  remain 
straight,  and  if  P  denotes  the  applied  force,  and  r  the  perpendicular 
distance  of  its  point  of  application  from  the  axis  of  the  shaft,  then 
'Pr  is  the  measure  of  the  twisting  moment.     In  the  case  of  the 
steam  engine,  P  varies  with  the  ratio  of  expansion  of  the  steam  in 
the  cylinder,  and  the  shaft,  in  addition  to  its  twisting  moment,  is 
subjected  to  a  slight  bending  moment  due  to  its  own  weight.     If 
the  twisting  moment  be  calculated  from  the  I.  H.  P.,  that  is,  from 
the  mean  pressure  in  the  cylinder,  it  is  known  as  the  mean  twisting 
moment.     Since  shafts  must  be  strong  enough  to  resist  the  maxi- 
mum stress  to  which  they  may  be  subjected,  the  moment  expressing 
the  maximum  stress  due  to  the  combined  twisting  and  bending 
actions  should  form  the  basis  of  the  calculation.     This  moment  is 
known  as  the  maximum  equivalent  twisting  moment.     The  ratio 
between  the  maximum  equivalent  twisting  moment  and  the  mean 
twisting  moment  varies  somewhat  with  different  kinds  of  engines, 
but  it  is  not  far  wrong  to  take  it  as  1.5. 

68.  It  has  been  shown  that  the  strength  of  a  section  to  resist 
bending  is  measured  by  its  modulus,  Z,  and  the  relation  -5-  =  -  —  Z 

has  been  shown.  By  reasoning  analogous  to  that  on  pages  313  and 
314,  it  may  be  shown  that  the  strength  of  a  section  to  resist  torsion 
is  measured  by  its  polar  modulus,  Zp,  and  we  shall  have  : 

£=£=*-•  (E> 

69.  In  the  case  of  a  solid  circular  shaft  for  transmitting  power, 
we  have  It  =  ?g,  and  c  =  |.  Hence  Z,  =  *j£  -s-  f  =  yf  =  0.1960. 

For  a  hollow  circular  shaft,  we  shall  have  : 


p~        32         '    2  " 

70.  If,  in  the  hollow  shaft,  we  let  -  denote  the  fraction  of  the 
solid  section  removed  for  the  axial  hole,  we  shall  have  : 


COLUMNS. — SHAFTS  325 

Solid  Area 
Eemoved  Area  = 


n 

That  is,  we  shall  have  ~  =  ^,  whence  d*  =  ~. 
Then  the  modulus  for  the  hollow  shaft  becomes : 


(P) 

It  is  the  common  practice  for  hollow  shafts  to  make  the  internal 
diameter  equal  to  one-half  the  external  diameter,  in  which  case  the 
part  removed  is  one-fourth  the  whole  area  of  the  section.  We  shall 

then  have  n  =  4,  and  Zp  =  ^-  (1  —  iV)  —    9gg    =  0.1 84D3. 


71.  Horse-power  Transmitted  by  Shafts.  —  Let  P  be  the  average 
force  in  pounds  acting  at  a  distance  r  inches  from  the  axis  of  the 
shaft.  Then  Pr  =  the  mean  twisting  moment  in  pounds-inches. 

If  N  be  the  number  of  revolutions  per  minute,  we  shall  have: 

,         2PmN 

Work  per  minute  in  foot-pounds  =  —  ^  —  . 

Then,  horse-power  transmitted  =  I.  H.  P.  =  12  x  33  000  ' 
whence  Pr  =  12  X  ^'^jf  L  H"  -  .    From  equation  (E)  we  have 

Pr  =  Mt  —  —  *  =  ~SZP  —  0.196&D3  =  Resisting  moment  for  a 
c 


solid  shaft.     From  equation  (F),  we  have  Pr  =  —  -r-l  ~~-^ 


=  0.  184:8  'Ds,  when  the  internal  diameter  is  one-half  the  external, 
and  n  —  4.  8  may  be  taken  as  7000  Ibs.  per  square  inch  for 
wrought  iron,  and  10,000  for  steel.  Taking  the  maximum  equiva- 
lent twisting  moment  as  1.5  times  the  mean  twisting  moment,  we 
shall  have  : 

0.196  x  70007>   =  13X3S.OOOXLH.P.  x  1.5     whMceB=  4- 


for  solid  wrought  iron  shafts. 

For  hollow  wrought  iron  shafts  with  inside  diameter  half  the 
outside,  we  shall  have : 

0.184  x  70002)3  =  1*  ><  88,000  x  I.  H.  P.  x  1.5 


326      THE  ELEMENTS  OF  THE  MECHANICS  OF  MATERIALS 

In  like  manner  we  shall  have  for  steel : 

D  =  '6. 64 A/  '    ,j      for  solid  shafts. 


a«d  3  p 

D  _  3.79   71-  **••  r-  for  hollow  shaft  having  external 
V       N  diameter  double  the  internal. 

EXAMPLE  V.— A  solid  steel  shaft  is  to  transmit  4000  I.  H.  P. 
while  making  130  revolutions  per  minute,  the  maximum  equivalent 
twisting  moment  being  1.5  times  the  mean;  find  its  diameter. 


D  =  3.6  =  3.64=10.13  inches. 


EXAMPLE  VI.  A  steel  shaft  is  to  have  30  per  cent  of  its  section 
removed  in  making  an  axial  hole,  and  is  then  to  transmit  9000 
I.  H.  P.  at  120  revolutions  per  minute.  Find  the  outside  and  inside 
diameters. 

Let  D  and  d  denote  the  outside  and  inside  diameters,  respectively. 
Then  the  area  of  the  removed  section  =  0.7854Z)2  X  0.3. 
And  d  =  -v/OaD'  =  0.5477/X 

We  have,  for  hollow  shafts,  Zp  =  ^-  (l  —  -^\  ,  iu  which  -  =  frac- 
tional area  removed.  Since  the  fractional  part  removed  is  30  per 
cent,  we  shall  have  -  =  ^,  whence  n=-^.  Therefore,  Zp 

7~k3 

=  ^  (1  —  Tthr)  =  .1787D3,  and  the  moment  of  resistance  =  SZP 
=  twisting  moment. 


Hence  0.178  W  X  10,000  =  12  X  ^'00  X  ^  ,  whence 


d  —  0.5477D  =  0.5477  X  15.83  =.  8.67  inches. 

EXAMPLE  VII.  —  Find  the  inside  and  outside  diameters  of  a  hol- 
low steel  shaft  to  transmit  8000  I.  H.  P.  at  130  revolutions  per  min- 
ute, the  unit  stress  in  the  outermost  fibers  not  to  exceed  10,000  Ibs. 
per  square  inch,  and  the  interior  diameter  to  be  98  per  cent  of  one- 
half  the  exterior  diameter.  Find  also  the  diameter  of  a  solid  shaft 
to  transmit  the  same  power  under  the  same  conditions.  Show  that 
the  two  shafts  are  equal  in  strength,  and  that  the  hollow  shaft  is 
20.87  per  cent  the  lighter. 


COLUMNS.  —  SHAFTS  327 

Let  D  and  d  denote  the  outside  and  inside  diameters,  respectively. 


.7854(.49Z))2  _    6 
Fractional  section  area  removed  by  axial  hole  =       7354/^2        -  25 

Hence  -  =  _  -^  ,  whence  n  =  ~  . 
n      25  b 


The  resisting  moment,  8ZP,  is  then,  10,000  X  0.185P3. 
The  maximum  equivalent  twisting  moment 

-  -  p     _  12  X  33,000  X8000  X  1.5 
27TX130 

Then,  10,000  X  0.186Z>-  =  18X88,000X^00X1.6  ,  whence 


n  _  7  12  X  33,000  X  8000  X  1.5   _  u  65  [     fa 

-V  2*  X  130  X  10,000  X  0.185  - 
Then,  d  =  14.65  X  0.49  =  7.19  inches. 


For  the  solid  shaft,  D  =  3.64  y  ^^'  =  3.64  y/^- 

=  14.37  inches. 

To  be  of  equal  strength  the  moduli  of  'the  sections  must  be  equal  ; 
that  is,  we  must  have  0.185  (14.65)  3  ==  0.196  (14.37)  3. 

The  weights  of  the  shafts  are  proportional  to  their  sectional  areas. 
Sectional  area  of  hollow  shaft  —  0.7854  [(  14.65  )2  —  (7.19)2] 
=  128.36  square  inches. 

Sectional  area  of  solid  shaft  =  0.7854(14.37)  2  =.  162.21  sq.  in. 

Hence  the  hollow  shaft  is  C16^1  —  128.36)100  _  2Q  g?         cent 

lo2.Icl 

lighter  than  the  solid  shaft. 

PKOBLEMS. 

58.  A  hollow  cylindrical  cast  iron  column  with  fixed  ends,  whose 
thickness  of  metal  is  2  inches,  length  24  feet,  and  outside  diameter 
24  inches,  is  to  be  used  in  a  building.     What  safe  load  can  it 
sustain?  Ans.  1,141,000  Ibs. 

59.  Find  the  diameter  of  a  steel  shaft  to  transmit  130  I.  H.  P. 
at  300  revolutions  per  minute.     (This  being  a  small  shaft  it  will 
be  lacking  in  stiffness,  and  in  consequence  0.5  inch  should  be  added 
to  the  calculated  diameter.)  Ans.  3.25  inches. 


328      THE  ELEMENTS  OF  THE  MECHANICS  OF  MATERIALS 

60.  Find  the  outside  diameter  of  a  hollow  steel  shaft  to  transmit 
10,000  I.  H.  P.  at  125  revolutions  per  minute,  the  inside  diameter 
to  be  one-half  the  outside  diameter.  Ans.  16  inches. 

61.  Find  the  outside  and  inside  diameters  of  a  hollow  steel  shaft 
to  transmit  3000  I.  H.  P.  at  200  revolutions  per  minute,  the  inside 
diameter  to  be  54  per  cent  of  the  outside  diameter,  the  maximum 
equivalent  twisting  moment  to  be  taken  at  1.5  times  the  mean,  and 
the  unit  stress  in  the  outermost  fibers  not  to  exceed  10,000  Ibs.  per 
sq.  inch.     Find  also  the  diameter  of  a  solid  shaft  to  transmit  the 
same  power  under  the  same  conditions.     Prove  the  equality  of 
strength  of  the  shafts,  and  show  that  the  hollow  shaft  is  25  per  cent 
the  lighter. 

Ans.  9.25"  and  5"  for  hollow  shaft;  9"  for  solid  shaft. 


CHAPTER  VI. 
THE  DEFLECTION  OF  BEAMS. 

72.  Deflection  of  Beams. — The  value  of  the  radius  of  curvature  of 
any  plane  curve  of  abscissa  x,  ordinate  y,  and  length  L}  is  given  in 
the  differential  calculus  as 


Wy  dzy  dx.d^y 

~dx*  ~dx* 

since  the  deflection  is  very  small,  and  dx  is  dL  in  the  limit. 


Substituting  this  value  of  r  in  equation  (C),  Art.  54,  we  have, 
M  =  -  \  y  ,  which  is  the  form  of  the  equation  of  the  elastic  curve 

CL9& 

for  investigating  the  deflection  of  beams. 

In  any  particular  case  the  value  of  M  must  be  expressed  in  terms 
of  x,  and  then,  after  two  integrations,  the  deflection,  y,  will  be 
known  for  any  value  of  x. 

Thus  for  a  simple  beam,  Fig.  58,  loaded  with  W  at  the  middle,  we 

Wx 

have  M  =  B±x  =  —^-  for  any  point  between  the  left  support  and 

the  middle  of  the  beam. 


Then  =       .    Integrating  once,  we  have,          .= 


When  x  =  -^  we  shall  have  ^=0,  whence  C  =  --  -y^-.     Then 

_Wx*       WI? 

4          16    ' 


330      THE  ELEMENTS  OF  THE  MECHANICS  OF  MATERIALS 


Wx*  WL'2x 

Integrating  the  second  time,  we  get,  Ely  =  -^  -  — jg — f-  & • 

=  0  when  x  =  0,  .  • .  C'  =  0. 
WDx 


Then  Ely  = 


16 


The  deflection  is  a  maximum  when 


=      '  hence  y-  =  -        = 


4817' 

73.  In  the  case  of  a  cantilever,  Fig.  59,  loaded  with  W  at  the  free 

end,  we  shall  have,  M  =  —  Wx.     Hence,  -  ,  ^  —  —  Wx  . 

Eldy  _       Wx*n 
~dx~          '^-+6- 

||  =  0  when  x  =  L,  consequently  C  =  ^L.     Then, 
J™  _  W,  and  */,  =  J^?  _  J£  +  6".    We  shall 


have  y  =  0  when  x  -  0,  .-.  C'  =  0.     Then  y  = 
The  deflection  is  a  maximum  when  x  —  L,  hence, 

IF 

-.  ™ 


-*•). 


74.  The  expressions  for  the  maximum  deflections  of  the  different 
kinds  of  beams  of  uniform  section,  under  different  conditions  of  sup- 
port and  of  loading,  may  be  obtained  by  similar  processes. 

75.  Points  of  Inflection.  —  Cantilevers  fixed,  or  built  in,  at  the 
end,  bend  concave  downward,  while  simple  beams  fixed  at  the  ends, 
and  those  which  overhang  the  supports,  bend  with  a  combined  curva- 
ture of  concave  downward  and  upward  —  concave  downward  near 
the    supports    and    concave    upward    toward    the    middle.      The 
points  at  which  these  contrary  curvatures  separate,  and  at  which 
there  is  no  bending  whatever,  are  known  as  points  of  inflection. 
They  can  be  found  by  placing  the  general  expression  for  the  bending 
moment  equal  to  zero.     Thus,  in  Example  I,  Art.  39,  it  is  evident 


THE  DEFLECTION  OF  BEAMS 


331 


from  inspection  that  there  will  be  a  point  of  inflection  between  the 
left  support  and  W2,  and  one  between  the  right  support  and  W 2. 
The  expression  for  the  bending  moment  at  any  section  distant  x 
from  the  left  support,  and  between  that  support  and  W2  is  R±x 

aty 

—  TT1(a;  +  4)  —  ^(x-^-8)2.      Placing  this   expression   equal  to 

zero,  and  substituting  the  values  of  Rly  W1?  and  w,.  the  resulting 
quadratic  gives  a  value  of  4.368  for  x,  which  shows  that  there  is  a 
point  of  inflection  at  4.368  feet  to  the  right  of  the  left  support.  The 
expression  for  the  bending  moment  at  any  section  between  W2  and 
the  right  support,  and  distant  x^  from  the  left  support,  is  Rlx1 

-W,(x,  +  4)  -  -  W2(x,  --  12)  -  -  Ifo  +  8)2.     Placing  this 

equal  to  zero,  the  resulting  quadratic  gives  a  value  of  17.045  for  x^ 
showing  that  the  other  point  of  inflection  is  at  the  section  distant 
about  17  feet  from  the  left  support. 


F/'p.  6O. 


76.  Beams  with  Fixed  Ends. — In  the  case  of  a  beam  built  in,  or 
a  fixed  "  at  the  ends,  Fig.  60,  and  uniformly  loaded  with  w  pounds 
per  unit  of  length,  we  have  a  condition  not  heretofore  encountered. 

By  "  fixed  "  is  meant  that  the  parts  of  the  beam  built  in  are  con- 
strained to  remain  horizontal  while  the  beam  is  being  bent,  but  it  is 
understood  that  the  beam  is  free  endwise. 

The  reaction  of  the  wall  in  keeping  the  built-in  portion  of  the 
beam  horizontal  is  equivalent  to  the  action  of  the  couple,  P,  P, 
causing  an  unknown  bending  moment,  Pz,  at  the  support,  and  nega- 
tive because  of  its  concave-downward  tendency. 

The  bending  moment,  due  to  this  reaction,  at  any  section  between 
the  left  support  and  the  middle,  and  distant  x  from  the  left  sup- 
port, is, 

P(z-i-x)  —  Px  —  Pz. 


332      THE  ELEMENTS  OF  THE  MECHANICS  OP  MATERIALS 

Hence,  the  fixing  of  the  beam  at  the  ends  causes  a  constant  bend- 
ing moment  over  the  entire  length  of  the  beam  which  is  quite  inde- 
pendent of  that  caused  by  the  load,  but  in  the  opposite  direction. 

Taking  the  origin  at  the  support,  the  bending  moment  at  any 
section  is, 


Hence, 

fS- 

The  first  anti-z-derivative  is, 

«!*!( 

-|  =  0  when  x  —  0  and  when  x  =     ,  •'.  C  —  0. 

Letting  x  =  -^9  we  have  : 

w(L*      L3\      PzL  wU      ..*..!,      A- 

2  (  -g  --  24  )  ---  2~~    ~    ?  whence  Pz  =  -y~-  ,  which  is  the  bending 

moment  at  the  support. 

Substituting  the  value  of  Pz  in  (1),  and  letting  x  =  ~-,  we  have: 
M      wlU      L>\      wL?      wU  n  n       w.L2 

-iVT^Tr-Tr*  -ir^-^-^w 

Therefore,  the  bending  moment  at  the  middle  is  but  half  that  at 
the  support,  and  we  have  : 


12         12  * 

For  the  points  of  inflection  we  place  the  second  derivative  equal 
to  zero,  thus  : 

«  (Lx  -  X2)  _  "V  =  o,  or  X*  -  x2  =  ^,  whence  x  =  ^/1 

^  A-  * 

Integrating  the  second  time,  we  get  : 

w/WB8       a;4N\       Pzx*       „, 

^-"T""  ' 


i/  =  0  when  a;  =  0  .  •  .  C"  =  0  ;   and  i/  is  a  maximum  when  a;  =  ^-  , 
consequently, 

F7-         _  w  /i4  _   Z*  \  _  wi4  _  W£4  /  1  _      1      _   1  '\       wZ*  _  PTi» 

^M*  -  2  V58        193/        "96"          2~  V48       192       48/        384  ~    384"' 

and  for  the  maximum  deflection,  we  have: 


THE  DEFLECTION  OF  BEAMS  333 

77.  Continuous  Beams. — The  beams  heretofore  considered  have 
been  either  cantilevers  or  simple  beams  having  two  supports.     A 
beam  resting  on  more  than  two  supports  is  termed  a  continuous 
beam. 

The  chief  difficulties  in  the  treatment  of  continuous  beams  are 
the  determination  of  the  support  reactions  and  the  bending  moments 

or  T 

at  the  supports.     These  being  determined,  the  formula  M  =  - 

is  applicable  to  the  question  of  the  safe  loading  of  the  beam. 

The  treatment  of  continuous  beams  is  somewhat  simplified  by  the 
use  of  Clapeyron's  Theorem  of  Three  Moments,  and  is  fully  set 
forth  in  any  complete  treatise  on  the  Mechanics  of  Engineering. 

Only  the  special  case  of  two  spans  uniformly  loaded  will  here  be 
considered. 

78.  Beam  Resting  on  Three  Supports,  having  two  equal  spans 
and  loaded  with  w  pounds  per  unit  of  length,  Fig.  61. 

i^/ 


L- 


The  bending  moment  at  any  section  distant  x  from  the  middle 
support  is, 


Hence, 


=  0  when  x  =  Q.:C  =  Q. 

Lx* 


aJ      ,    „, 

4 


y  =  0  when  x  =  0,  .-.  C"  =  0. 
The  three  supports  being  on  the  same  level,  it  is  known  that  y  =  0 

when  x  =  •„-,  consequently, 


334      THE  ELEMENTS  OF  THE  MECHANICS  OF  MATERIALS 


R         __-4- 
1  \16      4:8)       2  \32      48  "*"  192 

wL    ,      A    ,   -.x       wL      ,        a  p        3wL      3W 
^  +  1)  =  =^,  whence^  nr'  STT< 


where  TF  is  the  whole  distributed  load. 

q  TJ7 

From   symmetry,   Rn  =  ^  =  -  £.      Then    #3  =  V",   since 

ID  o 

B,  +  tf  2  +  B3  =  W. 

Placing  the  second  derivative  equal  to  zero  to  find  the  points  of 
inflection,  we  have: 

L         \      w/D       T     ,      \      0 

*     iT  ~ 


=}-z,  whence  a;  =     . 


Hence,  the  points  of  inflection  are  distant  one-eighth  of  the 
length  of  the  beam  from  the  middle  support,  and  on  either  side  of 
that  support. 

To  find  the  bending  moment  over  the  middle  support,  let  x  =  0 
in  (1),  and  we  have: 

wU  _        wD       WL 


_ 
*3         32"     "  ~8~  32         32  ' 

The  greatest  bending  moment  between  the  supports  will  be  at  the 
point  where  the  shear  passes  through  zero.  We  find  this  point  by 
placing  the  first  ^-derivative  of  the  moment  equal  to  zero,  thus: 

dM          3L  ,L  A     ,  5Z     ,,,.,, 

^-  —  ~"Tg"+o"  —  x=    '  whence  x  =  jg-  ;  that  is,  the  maximum 

Q  J 

bending  moment  between  the  supports  is  at  the  section  distant  -j^ 

from  the  end  support.     To  find  its  value,  substitute  the  value  of  x 
in  (1),  and  we  have: 

M  _  SwLfL  _  5ZA  _  W(  L*  _  51?  . 

W      ^\4     "  16 

wU  ,„>       ftn    ,   9^ 
"(64-  '- 


16 

9  WL 
- 


This  result  is  numerically  less  than  that  found  for  the  bending 

(WT  \ 
—  -qK-  j?  hence  the  maximum  bend- 

ing moment  is  at  the  middle  support,  and  we  shall  have  for  safe 
loading  : 

81  _  WL 

c   "  ~  32   ' 


THE  DEFLECTION  OF  BEAMS 


335 


The  bending-moment  and  shear  diagrams  are  now  readily  drawn, 
as  shown  in  Fig.  62. 


79.  The  relative  strength  of  beams  may  be  obtained  by  a  com- 
parison of  their  maximum  bending  moments,  and  their  relative 
stiffness  by  a  comparison  of  their  maximum  deflections.  The 
results  of  such  comparisons  for  certain  beams  of  uniform  section  are 
given  in  the  table  which  follows. 


Kind  of  Beam  and 
Nature  of  Load. 

Maximum 
Bending 
Moment. 

Maximum 
Deflection. 

Relative 
Strength. 

Relative 
Stiffness. 

Cantilever,  load  at 

WL 

WL3 

1 

1 

free  end. 
Cantilever,  load 

WL 

3E1 
WL3 

2 

2i 

uniformly  distributed. 
Simple  beam,  load 

2 
WL 

8EI 
WL3 

4 

*3 

16 

at  middle. 

Simple  beam,  load 
uniformly  distributed. 

Beam  with  fixed 

4 

WL 

~8~ 

WL 

48EI 

5WL3 
184EI 

WL3 

8 
8 

25| 
64 

ends,  load  at  middle. 
Beam  with  fixed 

8 
WL 

192EI 
WL3 

12 

128 

ends,  uniform  load. 

12 

384EI 

336      THE  ELEMENTS  OF  THE  MECHANICS  OF  MATERIALS 

PEOBLEM. 

62.  A  simple  beam  30  ft.  long  weighs  20  Ibs.  per  foot,  and  over- 
hangs each  support  6  ft.  It  bears  a  superimposed  load  of  100  Ibs, 
per  ft.  and  a  load  of  1400  Ibs.  concentrated  at  a  point  3  ft.  to  the 
right  of  the  middle.  Construct  the  shear  diagram  to  a  linear  scale 
of  -J  inch  to  the  foot,  and  a  load  scale  of  1  inch  =  2000  Ibs.  Find 
the  bending  moment  at  the  dangerous  section,  and  the  distances  of 
the  points  of  inflection  from  the  left  support. 

Ans.  Mmax  =  7760  lbs.-ft.,  and  the  points  of  inflection  are 
1.48  ft.  and  16.9  ft.  from  the  left  support. 


CHAPTER  VII. 
RESILIENCE. 

80.  Resilience.  —  If  a  bar  be  placed  in  a  testing  machine  and  sub- 
jected to  a  gradually  increasing  load  in  a  direction  to  produce  a 
tensile  stress,  the  elongation  produced  will  be  proportional  to  the 
load,  provided  the  elastic  limit  of  the  material  be  not  exceeded. 

The  load,  or  external  force,  will  gradually  increase  from  0  to  P, 

p 
and  its  mean  value  will  be  -Q-.     If  y  denotes  the  elongation,  or  dis- 

placement, the  expression  for  the  work  done  is,  -g  . 

-D  *A       on         i_          Stress      S       &  8L 

By  Art.  50,  p.  311,  we  have:  ^^  =  -  =  E.:  y  =  ^  . 

L 

By  Art.  49,  p.  255,  we  have  P  =  AS,  so  that  the  work  done  is 


But  AL  is  the  volume  of  the  bar,  consequently  the  work  done  is 

proportional  to  the  volume  of  the  bar,  and  therefore  to  its  weight. 

Within  the  elastic  limit,  the  work  done  in  stretching  the  bar  is 

S* 
known  as  the  Resilience  of  the  bar,  and  the  ratio,  7™,  is  known  as 

the  Modulus,  or  Coefficient,  of  Resilience,  in  which  S  is  the  stress 
at  the  elastic  limit  of  the  material. 

It  is  thus  seen  that  the  resilience  of  a  bar  is  found  by  multiplying 
its  volume  by  the  coefficient  of  resilience. 

EXAMPLE  I.  —  A  steel  bar  10  feet  long  and  2  inches  in  diameter  is 
stretched  to  the  elastic  limit  ;  what  is  its  resilience  ? 

Solution:  The  elastic  limit  of  steel  in  tension  is  50,000  pounds 
per  square  inch,  and  the  coefficient  of  elasticity  30,000,000  pounds 
per  square  inch. 

Then,  Resilience  =  X  3.1416  X  10  =  1309  ft.-lbs. 


81.  Resilience  from  Sudden  Loads  or  Shocks.  —  If  a  bar  be  sub- 

jected to  a  sudden  load,  P,  that  is,  a  load  of  the  same  intensity 

throughout  the  period  of  the  elongation,  y,  occasioned  in  the  bar, 

the  external  work  is  Py.     The  elongation,  y,  produces  an  internal 

22 


338      THE  ELEMENTS  OF  THE  MECHANICS  OF  MATERIALS 

stress,  R,  which  increases  gradually  from.  0  to  R  with  a  mean  value 
of  -£.  The  internal  work  is  then  -J^,  and  since  this  must  equal  the 

external  work,  we  find  that  R  =  2P,  and,  therefore,  the  work  done 
on  the  bar  is  2Py.  Hence,  the  resilience  due  to  a  suddenly  applied 
load  is  just  four  times  as  much  as  that  due  to  the  same  load  applied 
gradually. 

82.  Resilience  of  Beams.  —  The  work  performed  in  bending  a 
beam  to  the  maximum  deflection  within  the  elastic  limit  is  the 
resilience  of  the  beam. 

If  the  load  be  concentrated  at  one  point,  the  resilience  is  the 
product  of  half  the  load  into  the  deflection. 

Thus,  if  P  denotes  the  load  concentrated  at  the  extremity  of  a 
cantilever  that  will  cause  the  maximum  deflection  within  the  elastic 

cr  T 

limit,  the  maximum  bending  moment  will  be  PL  =  —  •,  whence 

P       SI 

=  Lc' 

pj'A        &T  i 

The  maximum  deflection  is 


SI    SU       S*    LI      S2    K2     AJ 
Consequently,  Resilience  =  ^.  3^  =  O  '  3?  =  2#'  W  '  AL> 


in  which  AK2  is  substituted  for  I;  see  Art.  45,  p.  306. 

For  rectangular  sections,  c  =  ~  and  K2  —    .  =  ™-~  =  y-g,  so  that 


Then,  for  cantilevers  of  rectangular  section  and  with  concen- 
trated load  at  the  free  end,  we  have  : 


Resilience  =,.,  (B) 

or,  it  is  the  product  of  the  coefficient  of  resilience  and  one-ninth  the 
volume  of  the  beam. 

83.  With  a  simple  beam  having  a  concentrated  load  at  the  middle, 

~P  T          Sf  T 

the  maximum  bending  moment  is  -^-  =  —  -,  whence  P  = 

PL3        SLS 
The  maximum  deflection  is 


2SI      SL2        S2     LI 
Consequently,  Resilience  =  --  .  =        .        ,  which  is  the 


same  as  that  found  for  the  cantilever  with  load  at  extremity. 


KESILIENCE  339 

Hence,  for  simple  beam  with  load  concentrated  at  the  middle, 
we  have  : 


Resilience  =  ~  .  ^  .  (C) 

84.  If  the  load  be  distributed,  the  half  load  on  each  elemental 
length  of  the  beam  must  be  multiplied  by  the  corresponding  deflec- 
tions and  the  resilience  found  from  the  summation  of  these  products. 

85.  If  a  cantilever  be  loaded  with  w  pounds  per  linear  unit,  the 
elemental  load  is  wdx9  and  if  y  be  the  corresponding  deflection,  the 

elemental  external  work  is  — ^ — . 

The  bending  moment  for  any  section  distant  x  from  the  free 

n  .         wx2 
end  is ;5-. 

A 

Hence,  El  ^  —  —  <u^j- , 

Integrating  once,  we  have,  El  -f-  =  —  -y  -f-  C. 

ctx  o 

nr  o  wnen  x  ^^  LJ  .*.  C/  ^^ . 

dx  6 


y  =  0  when  x  =  0  .-.  C'  =  0. 

Then,  y  =  oT^jC—  x*  +  4ZAc),  and  we  shall  have: 

7/.2  f\Jj 

Resilience  =  -J™  I     (—  a;4  +  4L3^)rfa; 


wa    f      x* 
481/L      5 


Mmax  =         =      ,  whence  w  =         . 
Consequently,  Resilience  = 

7^ 

For  rectangular  sections,—^  =  i,  hence, 

c 

O2 

Resilience  =^.  (D) 

for  cantilevers  of  rectangular  section  and  uniform  load. 


340      THE  ELEMENTS  OF  THE  MECHANICS  OF  MATERIALS 

86.  For  a  simple  beam,  uniformly  loaded  with  w  pounds  per 
linear  unit,  the  elemental  load  is  wdx  and  the  elemental  external 

work  for  the  deflection,  y,  is  —  |  — 

The  bending  moment  at  any  section,  distant  x  from  the  left  sup- 

,    .    wLx      wx1      w  ,j          2, 
port,  is  -g-  —  -2  -  =  .g  (Lx  —  z2). 


Hence,  EI       =     (Lx  —  z2). 
ax        A 

Integrating  once,  we  have, 


y  z=  0  when  x  =  0  .  • .  C'  =  0. 

Then,  y  =  n^Lj-  (2Lx5  —  x*  —  L3x),  and  we  have, 

Eesilience  =  A^T   I    (%Ltf  —  x*  —  Dx^dx 


__ 
48^7      2        5        2      o          240^7 


,,     -o    .,.  L      S2     8    K*    AT 

Consequently,  Eesilience  -=  =       .      .  -      .AL. 


For  rectangular  sections,  —y-  =  J,  hence 
'  Resilience  =. 


87.  Comparing  equations  (B),  (C),  (D),  and  (E)  with  equation 
(A)  we  observe  that,  for  rectangular  sections,  the  resilience  of  a  bar 
in  tension  is  9  times  as  great  as  that  of  a  cantilever  loaded  at  its  free 
end,  9  times  as  great  as  that  of  a  simple  beam  loaded  at  the  middle, 
10  times  that  of  a  cantilever  uniformly  loaded,  and  5f  times  that  of 
a  simple  beam  uniformly  loaded. 

88.  Resilience  of  Shafts  under  Torsion.  —  If  a  shaft  be  twisted 
by  a  force  acting  at  right  angles  to  the  axis,  each  elemental  area  of 


EESILIENCE  341 

its   cross-section  is  subjected  to  a  shearing  stress  which  varies 
directly  as  the  distance  of  the  elemental  area  from  the  axis. 

The  stress,  R,  on  an  elemental  area,  a,  gradually  increases  from 

•p 
0  to  R,  its  mean  value  being   -_• . 

The  angle  through  which  the  longitudinal  elements  are  twisted 
varies  directly  as  their  distance  from  the  axis  (Fig.  63).  If  tf 
denotes  the  angle  through  .which  the  elements  distant  r  from  the 
axis  are  twisted,  then  the  displacement,  y,  of  an  elemental  area,  a, 
for  a  length,  dx,  is  dx  tan  & ;  or  since  ff  is  very  small,  we  shall  have 
y  —  dxtf. 

Within  the  elastic  limit  we  have : 

?•  5- 


Then,  Elemental  Work  =  ^ay  =  ^a  dxd'  =  .  (1) 

"  -v  &EJ 

If  8  denotes  the  unit  shearing  stress  at  the  remotest  elemental 
area,  distant  c  from  the  axis,  then  —  is  the  unit  shearing  stress  at 
a  unif  s  distance  from  the  axis,  and  —  is  the  unit  shearing  stress 
at  the  elemental  area  distant  r  from  the  axis.  Hence,  R  =  —  ; 

and  in  like  manner  7^  =  —  '  ,  ^  =  ^  ,  &c. 

c  c 

Substituting  in  (1),  we  have: 


Elemental  Work  =  ^  ~        '~'dx~        '  ~    '  Adx'  since 


2JJJ  '  ~<?  ' 

+  airl  +  &c-  =  IP  —  polar   moment   of   inertia,    and   Ip  =  AK2. 
Integrating  this  over  the  whole  length  of  the  shaft,  we  have: 

S*    K* 
Work  Performed  =  Eesilience  of  Torsion  =  ~-™.  -^  .  AL. 

„       .      ,  D       ,  ^        I      nD*       ^Z)2        Z>3 

For  circular  sections,  c=^  and  A  3  =  —  =  -—  -5  —  —  =  —  , 

so  that  ~  =  \. 


342      THE  ELEMENTS  OF  THE  MECHANICS  OF  MATERIALS 

Then,  for  shafts  of  circular  sections,  we  have : 
Resilience  of  Torsion  =  ^ .  —^-9  or,  it  is  the  product  of  the 
coefficient  of  resilience  and  one-fourth  the  volume  of  the  shaft. 

89.  The  determination  of  the  resilience  of  any  given  material  is 
regarded  as  the  best  measure  of  its  capacity  to  withstand  shocks. 

PROBLEMS. 

63.  What  work  is  required  to  subject  a  steel  bar  to  a  tensile 
stress  of  20,000  pounds  per  square  inch  200  times  a  minute,  the 
length  of  the  bar  being  10  feet  and  its  diameter  2  inches  ? 

Ans.  1.269  H.  P. 

64.  For  a  rolled  steel  beam,  symmetrical  about  the  neutral  axis, 
the  moment  of  inertia  is  72  inch  units.     The  beam  is  8  inches  deep, 
is  laid  across  an  opening  of  10  feet,  and  carries  an  evenly  distrib- 
uted load  of  9  tons.     Find  the  maximum  fiber  stress,  also  the  central 
deflection,  taking  E  as  26,000,000. 

Ans.  7.5  tons,  and  0.216  inch. 

65.  A  10-inch  Cambria  I  beam,  weighing  25  Ibs.  per  ft.  is  laid 
across  an  opening  of  16  ft.     In  addition  to  a  concentrated  load  of 
1000  Ibs.  at  the  middle  it  bears  a  uniformly  distributed  load  of 
14,000  Ibs.     Find  the  maximum  fiber  stress  and  the  maximum  de- 
flection, taking  E  as  29,000,000.  Ans.  16,130  Ibs.,  0.416  inch. 

66.  A  beam  of  wood,  8  inches  wide  and  12  inches  deep,  and  fixed 
at  the  ends,  covers  a  span  of  14  ft.,  and  bears,  in  addition  to  a  con- 
centrated load  of  6000  Ibs.  at  the  middle,  a  uniformly  distributed 
load  of  5000  Ibs.     Find  the  maximum  fiber  stress  and  the  maximum 
deflection,  taking  E  as  1,500,000.          Ans.  1020  Ibs.,  0.128  inch. 


PART  V 
ELEMENTS  OF  GRAPHIC  STATICS 


CHAPTEE  I. 

BOW'S  SYSTEM  OF  LETTERING.     FORCE  DIAGRAM. 
FUNICULAR  POLYGON. 

1.  Graphic  Statics. — The  methods  by  which  static  problems  are 
solved  by  means  of  scale  drawings  constitute  Graphic  Statics. 

In  very  many  cases,  the  determination  by  calculation  of  the 
forces  transmitted  through  the  different  parts  of  a  structure  involve 
tedious  and  difficult  processes,,  with  the  consequent  liability  to  error, 
and,  in  the  end  the  effort  to  check  the  accuracy  of  the  results  occa- 
sions a  procedure  as  tedious  as  the  processes  themselves.  By  the 
graphic  method,  however,  solutions  are  readily  obtained,  and  the 
process  itself  furnishes  a  check  as  to  accuracy. 

2.  Bow's  System  of  Lettering. — To  the  method  of  lettering  dia- 
grams, devised  by  Mr.  A.  H.  Bow,  is  due  much  of  the  facility  in 
making  graphic  solutions. 

The  two  important  features  of  the  Bow  system  of  lettering  are: 
(1)  The  placing  of  a  letter  in  each  of  the  spaces  between  the  lines 
of  action  of  the  external  forces;  (2)  the  naming  in  clockwise  order 
of  each  force  by  the  two  letters  flanking  it. 

The  forces  of  Fig.  1  (a)  will  serve  to  illustrate  the  Bow  system. 
The  five  forces  are  in  equilibrium,  and  the  letters  A,  B,  C,  D,  and  E 
are  placed  in  the  spaces  between  them.  Other  letters,  or  even 
numbers,  would  serve  the  purpose,  and  they  might  be  placed  in  any 
order,  but  it  will  be  found  convenient  to  commence  at  the  left  and 
letter  the  spaces  alphabetically.  The  force  of  20  Ibs.  is  flanked 
by  the  letters  A  and  B  and  is  known  as  the  force  AB,  not  as  BA, 
since  the  letters  must  be  read  clockwise.  In  like  manner  the  re- 
maining forces  are  known  as  BC,  CD,  DE,  and  EA. 

3.  Force  Diagram. — Taking  the  forces  in  clockwise  order,  and 
denoting  them  by  the  corresponding  small  letters  of  the  alphabet, 
we  may  represent  them  in  magnitude  and  direction  in  a  diagram 
by  lines  drawn  to  some  chosen  scale.     Such  a  diagram  is  known  as 
a  force  diagram. 


346  ELEMENTS  OF  GRAPHIC  STATICS 

Thus,  A  being  the  first  letter  in  clockwise  order  from  the  left  in 
Fig.  1  (a),  the  letter  a  will  be  the  starting  point  of  the  force  dia- 
gram of  Fig.  1  (6),  and  since  the  force  AB  acts  downward,  ab,  to 
the  scale  of  20  pounds  to  the  inch,  will  represent  it.  The  force  BC, 
next  in  clockwise  order,  and  equal  to  40  Ibs.,  acts  upward  and  will 
be  denoted  by  be,  two  inches  in  length  and  measured  upward.  In 
like  manner,  cd,  measured  downward  and  one-half  inch  in  length, 
denotes  the  force,  CD,  of  10  Ibs.;  de,  measured  upward  and  one- 
quarter  inch  in  length,  denotes  the  force,  DE,  of  5  Ibs. ;  and,  finally, 
ea,  which  is  found  to  measure  three-quarters  of  an  inch,  properly 


/ 

20/6s 

?   \    B 

Tc 

tO/bs. 

\ 

\                & 

c 
e 
d 

L 

/&s. 

£ 
^/  o   / 

\ 

Slbs 

a 

i  •  y  •      • 

(a) 

(d) 

6 

denotes  the  force,  EA,  of  15  Ibs.  It  will  be  noted  that  the  force  EA 
is  of  just  sufficient  magnitude  to  fill,  when  drawn  to  scale,  the  space 
between  e  and  a  and  thus  close  the  force  diagram.  This  force  dia- 
gram is,  in  reality,  a  closed  polygon  which  has  resolved  itself  into 
a  straight  line  in  consequence  of  all  the  forces  being  vertical.  Had 
it  not  closed,  the  system  being  in  equilibrium,  there  would  have 
been  an  error  in  the  construction. 

With  a  correct  construction,  and  the  last  point  of  a  force  diagram 
not  falling  on  the  first  point,  would  indicate  a  resultant  force,  equal 
in  magnitude  to  the  scale  distance  between  the  last  point  and  the 
first  point,  and  acting  upward  or  downward  according  as  the  last 
point  had  fallen  above  or  below  the  first  point. 

4.  If  one  of  the  forces  of  Fig.  1  (a)  were  unknown  it  could  be 
found  by  means  of  the  force  diagram.  Suppose  the  force  BO  un- 


Bow's  SYSTEM  OF  LETTERING. — FORCE  DIAGRAM         347 

known.  Then,  commencing  with  the  force  CD,  the  force  diagram 
would  be  constructed  by  representing  in  clockwise  order  the  forces 
CD,  DE,  EA,  and  AB  by  the  scale  distances  cd,  de,  ea,  and  ab, 
respectively,  of  Fig.  1  (&),  c  being  the  first  point  of  the  diagram 
and  I  the  last  thus  determined.  The  missing  force  must  then  be 
represented  by  ~bc,  measured  upward,  and  as  it  measures  2  inches, 
it  correctly  does  so  for  the  force,  BC,  of  40  Ibs. 

5.  The  system  of  lettering  enables  the  resultant  of  any  number 
of  the  forces  to  be  read  at  once  from  the  force  diagram. 

Suppose  the  resultant  of  the  forces  EC,  CD,  and  DE  were  re- 
quired. The  first  and  last  letter  in  the  naming  of  these  forces  are 
B  and  E,  respectively,  so  that  be  on  the  force  diagram  represents 
the  required  resultant  in  magnitude  and  direction.  The  measure- 


ment  of  be  is  found  to  be  1.75  inches,  which,  to  the  scale,  repre- 
sents 35  Ibs.,  the  resultant  of  the  forces  40  +  5  — 10,  and  it  acts 
upward  since  be  is  measured  upward. 

6.  Should  the  system  of  forces  act  on  an  open  framed  structure, 
such  as  the  roof  truss  shown  in  Fig.  2,  a  letter  must  be  placed 
within  each  open  space  of  the  frame  in  addition  to  those  placed  in 
the  spaces  between  the  external  forces.  The  external  forces,  by 
Bow's  notation,  are  known  as  A  B,  BC,  CD,  DE,  and  EA.  The 
stress  in  the  member  connecting  joints  1  and  2  is  known  as  BG. 
The  stress  in  the  member  separating  the  spaces  lettered  F  and  G 
may  be  known  as  GF  or  as  FG  according  to  the  end  of  the  member 
under  consideration.  If  the  end  considered  is  that  at  joint  1,  then 
the  stress  in  the  member  is  known  as  GF,  because  the  forces  about 
that  joint,  taken  in  clockwise  order,  are  AB,  B G,  GF,  and  FA.  If 
the  other  end  of  the  member  is  under  consideration,  the  stress  is 
known  as  FG,  because  the  forces  about  the  joint  at  that  end,  in 
clockwise  order,  are  known  as  FG,  GH,  HI,  IE,  and  EF. 


348  ELEMENTS  OF  GRAPHIC  STATICS 

•7.  Funicular  Polygon. — Suppose  a  jointed  frame,  Fig.  3,  to  be 
acted  upon  at  its  hinged  joints  by  a  system  of  forces  in  equilibrium, 
the  members,  bars,  or  links  of  the  frame  being  free  to  adjust  them- 
selves to  the  best  position  for  withstanding  the  action  of  the  forces. 

Such  a  figure  is  called  a  funicular  polygon,  the  word  funicular 
having  no  mechanical  significance.  Since  the  system  is  in  equilib- 
rium, the  funicular  polygon  must,  of  course,  be  a  closed  polygon. 

The  equilibrium  is  occasioned  by  the  balance  between  the  internal 
forces,  or  stresses,  in  the  members  and  the  external  forces,  and  the 
whole  being  in  equilibrium,  each  joint  in  itself  is  in  equilibrium. 

Since  the  equilibrium  at  each  joint  is  the  result  of  the  action  of 
three  concurring  forces,  viz.,  the  external  force  at  the  joint  and  the 
stresses  set  up  in  the  two  members,  it  follows  that  a  triangle  may 


be  constructed  for  each  joint  which  will  represent  these  forces  in 
magnitude  and  direction. 

If,  for  example,  the  force  AB  be  known,  the  equilibrium  of  the 
joint  at  which  it  is  applied  is  maintained  by  the  action  of  the  force 
AB  and  the  stress  forces  in  the  members  AO  and  BO.  Draw  ab 
parallel  to  the  force  AB,  and  make  its  length,  to  some  chosen  scale, 
represent  the  magnitude  of  AB.  From  a  and  b  draw  lines  parallel 
to  AO  and  BO,  respectively,  intersecting  at  o.  Then  abo  is  the  tri- 
angle of  forces  for  the  joint  ABO,  and  bo  and  oa  represent  not  only 
the  directions  of  the  stresses  in  the  members  BO  and  OA,  but  their 
magnitudes  as  well,  to  the  same  scale  that  ab  represents  the  force 
AB.  The  stress  bo  in  the  end  of  the  member  BO  at  which  AB  is 
applied  occasions  an  equal  and  opposite  stress  ob  at  the  other  end 
of  BO,  and,  in  fact,  such  is  the  case  in  all  of  the  members  of  the 
polygon,  for  in  no  other  way  could  the  equilibrium  be  maintained. 
Taking  the  next  joint  in  clockwise  order,  we  have  the  external  force 
BC  and  the  stress  forces  CO  and  OB  in  equilibrium.  But  ob 


Bow's  SYSTEM  OF  LETTERING. — FORCE  DIAGRAM         349 

has  just  been  found  to  represent  in  magnitude  and  direction  the 
stress  OB.  Hence,  by  drawing  from  b  and  o  lines  parallel  respec- 
tively to  BC  and  CO,  intersecting  at  c,  we  shall  have  bco  as  the  tri- 
angle of  forces  for  the  joint  BCO,  and  be  and  co  will  represent  in 
magnitude  and  direction  the  force  BC  and  the  stress  in  the  member 
CO,  respectively.  We  now  know  the  stress  oc  at  the  joint  CDO, 
and  can  construct  the  triangle  of  forces,  cdo,  for  that  joint.  In 
like  manner  we  can  proceed  and  determine  all  the  external  forces 
and. stresses  in  the  members,  the  last  line,  fa,  closing  the  diagram, 
thus  proving  that  the  system  is  in  equilibrium. 

8.  It  will  be  observed  that  the  sides  of  the  polygon  just  con- 
structed represent  the  external  forces  of  the  funicular  polygon, 
and  therefore  abcdef  is  called  the  force  diagram.     Furthermore, 
all  the  lines  representing  the  stresses  in  the  members  of  the  funi- 
cular polygon  meet  at  a  point  called  the  pole.     These  stress  lines 
are  known  as  vectors. 

9.  From  what  has  preceded,  it  is  seen  that  if  all  the  external 
forces  that  are  applied  to  a  funicular  polygon  are  known  in  mag- 
nitude and  direction,  and  also  the  directions  of  two  of  its  members, 
the  force  polygon  can  be  drawn.     For,  the  force  diagram  can  be 
drawn  from  the  known  forces,  and  the  intersection  of  the  vectors 
parallel  to  the  known  directions  of  the  two  members  gives  the  pole. 
The  directions  of  the  remaining  members  are  then  found  by  draw- 
ing the  rest  of  the  vectors. 

10.  A  force  diagram  of  any  system  of  forces  in  equilibrium  can 
be  drawn,  and  by  choosing  any  pole,  a  funicular  polygon  with  respect 
to  .that  pole  can  then  be  constructed  to  which  the  forces  may  be 
applied. 

11.  The  closed  polygon  of  Fig.  4,  having  the  forces  AB,  BC,  CD, 
DE,  EF,  and  FA  applied  to  its  joints,  has  been  drawn  -to  illustrate 
some  of  the  properties  of  the  funicular  polygon.     Draw  the  force 
diagram  abcdef.     The  triangle  of  forces  abo  for  the  joint  ABO  de- 
termines the  pole,  and  the  other  vectors  may  then  be  drawn. 

Let  yy  be  any  section  of  the  polygon.  The  forces  to  the  right 
of  the  section  are  BC  and  CD  and,  by  the  triangle  of  forces,  their 
resultant  is  bd,  acting  from  b  to  d.  The  forces  on  the  left  of  the 
section  are  DE,  EF,  FA,  and  AB,  and,  by  the  polygon  of  forces, 
their  resultant  is  db,  acting  from  d  to  b.  Hence,  the  resultant  6f 


F  THE 


350 


ELEMENTS  OF  GKAPHIC  STATICS 


the  forces  on  one  side  of  a  section  has  the  same  magnitude  and  line 
of  action  as  the  resultant  of  the  forces  on  the  other  side,  but  acting 
in  opposite  directions  to  preserve  equilibrium. 

To  find  where  this  resultant  acts,  we  replace  the  forces  BC  and 
CD  by  their  resultant  Id,  so  that  our  force  diagram  now  becomes 
bdefa.  The  vectors  od,  oe,  of,  oa,  and  ob  have  o  as  their  pole, 
therefore  a  funicular  polygon  can  be  drawn  having  its  sides  par- 
allel to  these  vectors.  We  have  already  OD,  OE,  OF,  OA,  and  OB 
parallel,  respectively,  to  these  vectors;  but  since  a  funicular  poly- 
gon must  close,  and  since  there  must  be  a. force  acting  at  each  joint, 
we  produce  OB  and  OD  until  they  intersect,  and  at  the  joint  thus 
formed  the  resultant  bd  must  act. 


12.  An  inspection  of  the  funicular  polygon  and  the  force  dia- 
gram shows: 

(a)  The  resultant  of  the  forces  DE,  EF,  FA,  and  AB  to  the  left 
of  the  section  is  given  in  the  force  diagram  by  db,  the  first  and  last 
letters  of  the  forces  when  named  in  clockwise  order;   in  like  man- 
ner the  resultant  of  the  forces  BC  and  CD  to  the  right  of  the  sec- 
tion is  bd. 

(b)  The  letters  in  the  force  diagram  which  give  the  resultant 
also  name  the  members  of  the  funicular  polygon  which  have  to  be 
produced  to  intersection  in  order  to  get  a  point  in  the  line  of  action 
of  the  resultant. 

Thus,  the  resultant  of  the  forces  DE,  EF,  and  FA  is  da  in  mag- 
nitude and  direction,  and  producing  the  members  D  and  A  of  the 
funicular  polygon  to  their  point  of  intersection  we  get  a  point  in 
the  line  of  action  of  this  resultant. 


Bow's  SYSTEM  OF  LETTERING. — FORCE  DIAGRAM         351 

13.  To  make  a  practical  application  of  the  funicular  polygon, 
take  the  beam  of  Fig.  22,  page  289,,  reproduced  in  Fig.  5.  The 
linear  scale  is  1"  =  4',  and  the  load  scale  is  1"  =150  Ibs.,  so  that 
Wi  =  60  Ibs.  =  T6-gV  —  0.4  inch  to  scale,  W2  =  45  Ibs.  =  T4^  =  0.3 
inch,  and  Ws  =  90  Ibs.  =  -ff^  =  0.6  inch.  Letter  the  beam  ac- 


cording  to  the  Bow  system.  Then,  Wly  W2,  W3,  R2,  and  R±  will 
be  known  as  AB,  BC,  CD,  DE,  and  EA,  respectively. 

To  construct  the  force  diagram,  we  set  off  vertically  down- 
ward, ab  equal  to  0.4  inch  to  represent  W19  or  AB,  to  scale;  ~bc 
equal  to  0.3  inch  to  represent  W2,  or  BC ' ;  and  cd  equal  0.6  inch  to 
represent  W3,  or  CD.  Then  we  know  that  da  is  the  closing  line 
of  the  force  diagram,  all  the  forces  being  vertical,  and  that  it  rep- 
resents the  sum  of  the  reactions  R2  and  R19  but  we  do  not  know  the 
proportion  of  the  load  borne  by  each  support. 

To  find  R!  and  R2,  we  must  construct  the  funicular  polygon. 


352  ELEMENTS  OF  GRAPHIC  STATICS 

Select  at  random  some  point  o,  distant  oh  from  ad,  and  draw  the 
vectors  oa,  ob,  oc,  and  od.  From  some  point  j  in  the  line  of  action 
of  R19  draw  a  line  parallel  to  the  vector  ao,  and  from  its  point  of 
intersection,  k,  with  the  vertical  from  W±  draw  a  line  parallel  to 
the  vector  bo,  and  from  its  point  of  intersection,  f,  with  the  line  of 
action  of  W2  draw  a  line  parallel  to  vector  co,  and  from  its  point 
of  intersection,  ra,  with  the  line  of  action  of  Ws  draw  a  line  par- 
allel to  the  vector  do.  This  last  line  intersects  the  line  of  action 
of  R2  at  n.  Join  n  with  /,  and  we  have  the  funicular  polygon 
jkfmnj,  with  the  external  forces  of  the  beam  acting  at  its  joints. 
This  funicular  polygon  has  five  sides,  while  there  are  but  four  vec- 
tors in  the  force  polygon,  and  since  there  must  be  a  vector  for  each 
side  of  the  funicular  polygon,  the  vector  oe,  parallel  to  nj,  must 
be  drawn.  The  point  e  is  thus  determined,  and  de  represents  in 
magnitude  and  direction  the  support  reaction  DE  =-R2  to  the  scale 
adopted ;  and  in  like  manner  ea  represents  in  magnitude  and  direc- 
tion the  support  reaction  EA  =  R^.  By  measurement  de  is  |-J- 
inch,  which,  reduced  to  scale,  equals  ££  X  150  =  127.5  Ibs.  =  R2 ; 
and  ca  measures  f-J,  which,  to  scale,  equals  fj  X  150  =67.5  Ibs. 
=  RI.  These  are  the  same  values  found  before  for  R±  and  R2. 

It  will  not  be  necessary  to  letter  the  funicular  polygon  in  order 
to  know  the  names  of  the  members,  as  an  examination  of  the  force 
polygon  at  once  discloses  them.  Thus,  the  member  parallel  to  ao 
is  AO,  the  one  parallel  to  bo  is  BO,  and  so  on.  The  names  of  the 
members  may  also  be  determined  from  the  lettering  of  the  beam, 
since  the  length  of  each  member  is  terminated. by  the  lines  of  action 
of  two  forces  of  the  beam.  For  example,  the  member  Jef  is  ter- 
minated by  the  lines  of  action  of  W±  and  W2,  and  since  the  letter 
B  appears  between  W±  and  Wz,  the  member  kf  is  named  BO. 

14.  The  Funicular  Polygon  a  Bending-Moment  Diagram. — Con- 
sider any  section  x  of  the  beam  of  Fig.  5.  Produce  the  members 
jk  and  kf  until  they  intersect  the  vertical  through  x  at  p  and  q, 
respectively.  Draw  the  horizontal  lines  rs,  tu,  and  vw,  and  regard 
them  as  the  altitudes  of  the  triangles  jpz,  kpq,  and  fqy,  respectively. 
All  the  triangles  of  the  force  polygon  have  the  same  altitude  oh. 
The  triangles  jpz  and  oea  are  similar,  having  their  sides  mutually 
parallel.  For  the  same  reason  the  triangle  kpq  is  similar  to  the 
triangle  oba,  and  triangle  fqy  is  similar  to  the  triangle  ocb.  Hence, 
we  shall  have : 


Bow's  SYSTEM  OF  LETTERING. — FORCE  DIAGRAM         353 

ffi  =  «*  =  7^'  whence  RiXrs^pzXoh; 
oJi=^l  =  W'  whence  wiXtu  =  pqX  oli ;   and 
^=  |^  =  |r,  whence  F2  X  tw  =  ^  X  oft. 
The  bending  moment,  M,  at  section  x,  is, 

M  =  R^X  rs—W1  Xtu—W2Xvw 

—  pzX  oh—pq  X  oh  —  qyXoh  —  oh[pz  —  (pq  +  gy)] 

=  o/t  X  2/2- 

Hence  the  bending  moment  at  any  section  of  the  beam  is  equal 
to  the  product  of  the  polar  distance,  oh,  and  the  ordinate  of  the 
funicular  polygon  at  the  section.  Thus,  the  ordinate  g  measures 
|-{j-  inch,  and  oh  measures  1  inch  =  150  Ibs.  to  scale;  the  bending 
moment  at  the  beam  section  above  g  is  then,  150  Ibs.  X  IT  X  4  ft. 
=  217.5  lbs.-ft,,  the  same  as  was  found  on  page  290. 

15.  The  forces  W19  W2,  and  W3  are,  by  the  Bow  system,  known 
as  AB,  BC,  and  CD,  and  their  resultant,  ad,  acts  through  i  (see 
Art,  11). 

16.  The  shear  diagram  can  readily  be  constructed  by  projection 
from  the  force  diagram. 

The  shear  at  any  section  between  the  left  support  and  W±  is 
R^  =  ea.  This  shear  is  plotted  by  projecting  ea  horizontally,  as 
shown.  The  shear  at  any  section  between  W^  and  W2  is  R^  —  W± 
=  ea  —  ab  =  eb,  and  this  shear  is  plotted  by  projecting  e~b.  The 
shear  at  any  section  between  W2  and  W3  is  R±  —  Wt  —  W 2  =  ea 
-  ab  —  ~bc  —  —  ec,  and  the  shear  at  any  section  between  W3  and 
the  right  support  is  R1  —  W^  —  W2  —  W3  =  ea  —  ab  —  be  —  cd 
=  —  ed  =  —  R2.  Projecting  these  two  shears,  the  diagram  is 
completed. 

17.  It  has  been  stated  that  the  pole,  o,  may  be  selected  at  ran- 
dom, but  if  it  be  so  selected  that  its  distance  from  the  load  line  be 
some  decimal  multiple  of  pounds  or  of  tons,  as  the  case  may  be, 
then  a  bending  moment  scale  may  readily  be  obtained  that  will 
enable  the  bending  moment  to  be  measured  direct  from  the  ordinate 
of  the  diagram,  thus  saving  the  necessity  of  multiplying  each  meas- 
urement by  the  polar  distance.     The  pole,  o,  of  Fig.  5  was  taken  at 
a  distance  of  1.  inch  from  the  load  line,  ab,  the  polar  distance,  oh, 


354 


ELEMENTS  OF  GRAPHIC  STATICS 


therefore  representing  150  Ibs.     Then,  since  the  linear  scale  is 
1"  —  4',  the  bending  moment  scale  will  be  I"  —  4'  X  150  Ibs.  =  600 

21  75 
Ibs.-feet,  or  -fa  inch  =  10  Ibs.-feet.     The  ordinate  g  measures     -- 


f—     8 
\          A 


inch,  and  the  bending  moment  at  that  section  of  the  beam  is  217.5 
Ibs.-feet. 

EXAMPLE  I. — A  beam,  supported  at  the  ends,  is  20  feet  long, 
weighs  400  Ibs.,  and  has  concentrated  loads  of  360  Ibs.  and  440  Ibs. 
at  8  feet  from  the  left  end  and  4  feet  from  the  right  end,  respec- 


Bow's  SYSTEM  OF  LETTERING. — FORCE  DIAGRAM         355 

tively.  Draw  the  bending-moment  and  shear  diagrams,  and  meas- 
ure the  bending  moments  under  the  concentrated  loads,  and  the 
shear  stress  at  the  middle  of  the  beam. 

Select  a  linear  scale  of  -J  inch  =  1  foot,  and  a  load  scale  of 
1  inch  =  400  Ibs. 

Set  out  the  beam  as  shown  in  Fig.  6,  and  since  the  beam  weighs 
400  Ibs.  it  has,  in  addition  to  the  concentrated  loads,  a  uniformly 
distributed  load  of  20  Ibs.  per  foot. 

We  shall  first  construct  the  funicular  polygon  for  the  concen- 
trated loads,  neglecting,  for  the  present,  the  uniformly  distributed 
load. 

Set  off  the  load  line  ac  by  making  ab  =  f  $£  =  0.9  inch,  and 
be  =  |-£$  =  l.l  inches.  Select  the  pole,  o,  at  a  distance  of  1.25 
inches  from  ac,  so  that  the  polar  distance  will  represent  1.25  X  400 
=  500  Ibs.  We  shall  then  have  for  the  bending-moment  scale, 
J  inch  =  1  ft.  X  500  Ibs.  =  500  lbs.-ft,  or  TV  inch  =  100  Ibs.-ft. 

Draw  the  vectors  ao,  ~bo,  and  co.  From  a  point  1  in  the  line  of 
action  of  R^  draw  a  parallel  to  ao  and  produce  it  until  it  intersects 
the  line  of  action  of  W±  at  the  point  2.  From  2  draw  a  parallel  to 
fro,  producing  it  until  it  intersects  the  line  of  action  of  W2  at  the 
point  3.  From  3  draw  a  line  parallel  to  co,  intersecting  the  line 
of  action  of  R2  at  the  point  4.  Join  4  with  1  as  the  closing  mem- 
ber of  the  funicular  polygon  1234.  Draw  od  parallel  to  4--1. 
Then  the  polygon  1234  is  the  diagram  of  the  bending  moments,  and 
cd  and  da  the  support  reactions  due  to  the  concentrated  loads,  Wl 
andF2. 

The  distributed  load  equals  20  X  20  =  400  Ibs.  =  £§-£  =  *  incn 
to  scale,  and  one-half  of  this  is  to  be  borne  by  each  support. 

The  bending-moment  diagram  for  the  distributed  load  is  of  para- 
bolic form,  and  will  be  placed  above  the  diagram  for  the  concen- 
trated loads,  so  that  the  sum  of  the  ordinates  of  the  two  diagrams 
under  any  section  of  the  beam  will  measure  the  bending  moment  at 
the  section. 

Under  the  supposition  that  the  whole  of  the  distributed  load  of 
400  Ibs.  is  concentrated  at  the  middle  point  of  the  beam,  the  funicu- 
lar polygon  for  this  load  must  be  constructed  on  the  closing  member 
4--1  of  the  polygon  1234.  To  do  so  we  will  use  the  same  pole,  o, 
that  was  used  for  the  funicular  polygon  of  the  concentrated  loads. 
Lay  off  de  and  df,  each  one-half  inch  in  length  to  represent  the 
halves  of  the  distributed  load.  Draw  the  vectors  oe  and  of.  From 


356  ELEMENTS  QF  GRAPHIC  STATICS 

1  and  4  draw  parallels  to  of  and  oe,  respectively,  intersecting  at  s. 
Then  s!4  is  the  funicular  polygon  for  the  distributed  load,  and  the 
ordinate  st  measures  the  bending  moment  at  the  middle  point  of 
the  beam  due  to  the  distributed  load,  supposing  it  all  to  be  concen- 
trated at  the  middle  point.  It  can  readily  be  shown  that  the  bend- 
ing moment  at  the  middle  of  a  simple  beam,  for  a  uniformly  dis- 
tributed load,  is  only  half  that  due  to  the  same  load  when  concen- 
trated at  the  middle.  Hence,  the  ordinate  5t  measures  the  bending 
moment  at  the  middle  of  the  beam  due  to  the  distributed  load,  the 
point  5  being  the  middle  point  of  st.  A  parabola  constructed  on  1--4 
as  a  chord,  and  passing  through  the  points  1,  5,  4,  is  the  bending 
moment  diagram  due  to  the  uniformly  distributed  load, 

The  complete  bending-moment  diagram  for  the  beam  is  12345. 

The  reaction,  R,,,  measured  on  the  load  line,  is  cd  -(-  df  —  f-J- 
inches;  hence,  R2  =  f-J  X  400  =  696  Ibs.  The  reaction  R^  =  da 
+  de  —  ||  inches;  hence,  R±  =  £}  X  400  =  504  Ibs. 

The  ordinates  y  and  y',  of  the  bending-moment  diagram,  under 
W±  and  W2  measure  f£  inch  and  f -§-  inch,  respectively.  The  bend- 
ing moments  under  W±  and  under  W2  are,  therefore,  3400  Ibs.-ft. 
and  2600  Ibs.-ft.,  respectively.  These  results  may  easily  be  checked 
by  calculation. 

The  shear  at  the  left  support  due  to  the  uniformly  distributed 
load  is  equal  to  half  that  load,  and  is  positive  because  of  its  clockwise 
iendency.  This  shear  gradually  decreases  from  left  to  right  until, 
.at  the  middle,  it  becomes  zero,  and  at  the  right  support  it  again 
becomes  equal  to  half  the  load,  but  negative  owing  to  its  contra- 
'dockwise  tendency.  The  diagram  d'e'f'd"  therefore  represents  the 
:shear  due  to  the  distributed  load. 

The  total  shear  at  the  left  support  is  R^  and  is  equal  to  da  +  de. 
'The  parallel  to  e'f  shows  the  gradual  decrease  of  the  shear  from 
the  left  support  to  Wl  due  to  the  uniform  load.  Passing  W^  the 
shear  suddenly  drops  to  n  and  becomes  negative,  m'n  being  equal 
to  db.  np  parallel  to  e'f  shows  the  gradual  increase  in  the  shear 
from  W±  to  W2  due  to  the  distributed  load.  Passing  W2  the  sheai 
suddenly  drops  to  q,  pq  being  equal  to  be.  qv  parallel  to  e'f  shows 
the  further  increase  in  the  shear  due  to  the  distributed  load  until, 
at  v,  it  becomes  —  R2  =  —  (dc  +  df).  The  ordinate,  uw,  at  the 
middle  of  the  beam  measures  -£$  inch ;  the  shear  at  the  middle  sec- 
tion is,  therefore,  ^  X  400  =  56  Ibs. 


Bow's  SYSTEM  OF  LETTERING. — FORCE  DIAGRAM:        357 


fe  /r 


18.  The  example  of  Fig.  43,  page  300,  of  a  beam  with  overhang- 
ing ends,  loaded  uniformly  with  20  Ibs.  per  foot  and  with  two  con- 
centrated loads,  may  readily  be  solved  by  means  of  the  funicular 
polygon. 

Set  off  the  beam  as  shown  in  Fig.  7.     Divide  the  beam  into  any 


358  ELEMENTS  OF  GRAPHIC  STATICS 

convenient  number  of  parts,  eight  in  this  instance,  so  that  each 
part  will  be  4  feet  in  length,  and  will  bear  80  Ibs.  of  the  distributed 
load.  These  eight  parts  constitute  as  many  external  forces,  each 
acting  at  its  center  of  gravity. 

Letter  the  beam  according  to  the  Bow  system,  and  adopt  the  fol- 
lowing scales :  Linear,  y1^-  inch  =  1  foot ;  load,  1  inch  =  400  Ibs. 
We  shall  then  have:  W^  =  £££  =  0.5  inch,  W2  =  £  £  £  =  1  inch, 
and  each  of  the  equal  divisions  will  be  -ffo  =  0.2  inch. 

Set  off  the  load  line,  dk,  accordingly.  Select  a  pole,  o,  distant 
1.5  inches  from  the  load  line,  so  that  the  polar  distance  will  repre- 
sent 1.5  X  400  =  600  Ibs.  We  shall  then  have :  TV  inch  =  1' 
X  600  Ibs.  =  600  Ibs.-f t.,  or  -fa  inch  =  100  Ibs.-ft.  for  the  bending- 
moment  scale. 

Draw  the  vectors  of  the  force  polygon.  From  some  point  m  in 
the  line  of  action  of  E^  draw  a  parallel  to  ao,  and  produce  it  until 
it  intersects  the  line  of  action  of  AB  at  n.  Through  n  draw  a 
parallel  to  &o,  producing  it  to  its  intersection,  p,  with  the  line  of 
action  of  BC,  or  of  W^.  Draw  parallels  to  the  other  vectors  as 
shown.  The  member  qr,  parallel  to  oj,  intersects  the  line  of  action 
of  JK  at  r.  The  parallel  to  ok  through  r  intersects  the  line  of 
action  of  R2  at  s.  The  closing  member  is  then  sm,  and  mnp-  --qrs 
is  the  funicular  polygon,  or  the  bending-moment  diagram  of  the 
beam. 

Draw  ol  parallel  to  sm.  Then,  la  and  Tel  represent  to  scale  the 
reactions  R^  and  R2,  respectively. 

It  should  be  noted  that  ordinates  within  mnpt  and  sur  give  nega- 
tive bending  moments,  and  that  t  and  u  are  the  points  of  inflection. 

The  ordinates  y  and  y'  give  the  maximum  negative  and  positive 

144  . 
bending  moments,  respectively.    These  ordinates  measure    ^    inch 

and  £$  inch,  so  that  the  bending  moments  are  1440  and  1600  Ibs.-ft. 
The  shear  of  the  left  overhang  is  negative  because  of  its  contra- 
clockwise  tendency.  Commencing  at  the  left  end  of  the  beam,  the 
shear  increases  from  zero  to  — ab  when  TF±  is  reached.  Passing 
TFj  the  shear  becomes  — ac,  and  increases  up  to  the  left  support, 
where  it  becomes  — ad.  Passing  the  left  support  it  becomes 
ft^  —  ad  =  Id  and  decreases  up  to  TF2  where  it  becomes  Id  —  dg 
=  Ig.  Passing  W2  it  becomes  Ig  —  gh  =  —  lh,  and  gradually  in- 
creases until  the  right  support  is  reached,  where  it  becomes 
—  lh  —  Jiz  —  —  lz  (there  are  but  one  and  one-half  divisions  of  the 


Bow's  SYSTEM  OF  LETTERING. — FORCE  DIAGRAM 


359 


beam  between  W2  and  the  right  support).  Passing  the  right  sup- 
port the  shear  again  becomes  positive  and  equals  R2  —  Iz  =  zk, 
and  then  decreases  to  zero  at  the  right  end  of  the  beam. 

19.  To  draw  the  bending-moment  and  shear  diagrams  of  the 
cantilever  with  concentrated  loads  as  shown  in  Fig.  8,  we  proceed  as 
follows : 

Draw  the  load  line  ad,  making  ab,  be,  and  cd  equal,  to  some 
selected  scale,  to  the  loads  AB,  EC,  and  CD,  respectively.  Select 


some  pole,  o,  and  draw  the  vectors  ao,  bo,  co,  and  do.  From  some 
point  m  in  the  support  line,  draw  a  parallel  to  ao,  producing  it  until 
it  intersects  the  line  of  action  of  AB  at  n.  From  n  draw  a  parallel 
to  bo  until  it  intersects  the  line  of  action  of  EG  at  p.  From  p  draw 
a  parallel  to  co  until  it  intersects  the  line  of  action  of  CD  at  q. 
From  q  draw  a  parallel  to  do  to  meet  the  line  of  support  at  r.  Then, 
mnpqr  is  the  bending-moment  diagram  of  the  cantilever,  and  the 
ordinate  under  any  section  of  the  beam  is  the  measure  of  the  bend- 
ing moment  at  the  section.  The  maximum  bending  moment  is, 
of  course,  at  the  wall. 

The  shear  diagram  is  drawn  by  projection  from  the  load  line  and 
presents  no  difficulties. 


360 


ELEMENTS  OF  GRAPHIC  STATICS 


20.  In  the  case  of  a  cantilever  it  is  found  convenient  to  select  the 
pole  at  some  chosen  perpendicular  distance  from  either  extremity  of 
the  load  line.     In  Fig.  8  the  pole,  o,  might  just  as  well  have  been 
chosen  at  o'  and  an  equal  bending-moment  diagram,  mrip'q'r',  con- 
structed, as  shown  by  the  dotted  lines. 

21.  The  example  of  Fig.  30,  page  295,  of  a  cantilever  with  a  com- 
bination  of   concentrated   and   uniformly   distributed   loads   may 
easily  be  solved  by  means  of  the  funicular  polygon. 

Using  the  same  linear  scale  of  -f^  inch  =  I  ft.  as  on  page  294, 


44lbs 


.  9. 


and  a  load  scale  of  400  Ibs.  to  the  inch,  we  sliall  first  construct  the 
funicular  polygon  -for  the  concentrated  loads. 

Reduced  to  scale,  AE  =  -fa  =  TW  inch,  and  EG  =  ^  =  Tfo 
inch.  Set  off  the  load  line  ac,  Fig.  9,  making  ab  and  be  equal  to 
yVg-  inch  and  T^  inch,  respectively. 

Choose  a  pole,  o,  at  a  perpendicular  distance  of  one  inch  from  cy 
so  that  the  polar  distance,  oc,  represents  400  Ibs.  We  shall  then 
have  for  the  bending-moment  scale,  -£$  inch  =  1  ft.  X  400  Ibs. 
=  400  lbs.-ft.,  whence  -gV  inch  —  40  Ibs.-ft.  Draw  the  vectors  ao, 
bo,  and  co. 

Commencing  at  some  point,  m,  in  the  wall  line,  draw  parallels 
to  ao,  bo,  and  co  in  the  usual  manner,  thus  forming  the  funicular 


Bow's  SYSTEM  OF  LETTERING. — FORCE  DIAGRAM         361 

polygon,  or  bending-moment  diagram,  mnpq,  for  the  concentrated 
loads. 

The  cantilever  is  uniformly  loaded  with  48  Ibs.  per  foot  for  a 
distance  of  4  feet  from  the  wall,  making  a  total  uniform  load  of 
192  Ibs.  Assuming  this  load  concentrated  at  the  outer  extremity 
of  the  4  feet,  set  off  cd  equal  to  %$ f  =  f£  incn  to  represent  it. 
Draw  the  vector  do,  and  from  r,  the  intersection  of  pq  with  the 
vertical  at  4  feet  from  the  wall,  draw  rt  parallel  to  do.  Since  the 
bending  moment  due  to  a  concentrated  load  at  the  end  of  a  canti- 
lever is  twice  that  due  to  the  same  load  uniformly  distributed,  the 
distance,  qt,  must  be  bisected  at  s,  and  the  parabola  having  its  apex 
at  r,  and  passing  through  s,  gives  rsq  as  the  bending-moment  dia- 
gram due  to  the  distributed  load.  Then,  mnprs  is  the  complete 
bending-moment  diagram  for  the  cantilever.  The  ordinate,  ms, 

25  3 

at  the  wall  measures  "-RTT-  inch,  and  the  bending  moment  is,  there- 
fore, 25.3  X  40  =  1012  lbs.-ft.,  the  same  as  found  on  page  294. 

The  shear  diagram  is  drawn  by  projection  from  the  load  line; 
and  presents  no  difficulties. 

PEOBLEMS. 

1.  A  beam,  22  feet  long,  supports  a  load  of  1000  Ibs.  at  a  point 
6  feet  from  the  left  end  and  one  of  1200  Ibs.  at  5  feet  from  the 
right  end.     In  addition,  there  is  a  uniformly  distributed  load  of 
100  Ibs.  per  foot.     Construct  the  shear  and  bending-moment  dia- 
grams.    What  concentrated  load  must  be  placed  at  the  middle  of  a 
similar  beam  in  order  that  the  maximum  B.  M.  shall  be  the  same  as 
that  of  the  given  beam?  Ans.  2190  Ibs. 

2.  A  beam  30  feet  long  weighs  20  Ibs.  per  foot  and  overhangs 
each  support  6  feet.     It  bears  a  superimposed  load  of  100  Ibs.  per 
foot,  and  a  load  of  1400  Ibs.  concentrated  at  a  point  3  feet  from 
the  right  of  the  middle.     Construct  the  bending-moment  and  shear 
diagrams,  and  find  graphically  the  bending  moment  at  the  danger- 
ous section  and  the  distances  of  the  points  of  inflection  from  the 
left  support.  Ans.  7760  lbs.-ft;   1.48  ft.  and  16.9  ft. 

3.  A  cantilever,  14  ft.  long,  supports  three  concentrated  loads — 
500  Ibs.  at  4  ft.  from  the  wall,  600  Ibs.  at  10  ft.  from  the  wall,  and 
200  Ibs.  at  the  extremity.     In  addition,  it  bears  a  uniformly  dis- 
tributed load  of  80  Ibs.  per  foot  run.     Construct  the  bending-mo- 
ment and  shear  diagrams. 


CHAPTEE  II. 
FRAMED  STRUCTURES.     RECIPROCAL  DIAGRAM. 

22.  Framed  Structures. — A  framed  structure  is  an  assemblage  of 
members  for  the  transmission  or  modification  of  external  forces, 
the  internal  stresses  occasioned  thereby  in  the  members  being  prin- 
cipally those  of  tension  and  compression.     A  member  in  tension  is 
known  as  a  tie ;  if  in  compression,  it  is  known  as  a  strut. 

23.  A  "  frame  "  is  a  theoretical  structure,  the  joints  connecting 
its  members  being  supposed  frictionless.     There  are,  of  course,  no 
such  things  as  frames,  since  all  joints  offer  some  resistance  to  rota- 
tion.     In   general,   however,   engineering   structures   approach   so 
nearly  to  frames  that  no  sensible  error  results  from  treating  them 
as  such. 

24.  The  load  on  a  structure  consists  of  the  weight  of  the  struc- 
ture itself  and  the  external  forces  acting  upon  it.     Wind  pressure, 
and  the  weight  of  the  covering  of  the  structure  and  of  the  snow  it 
may  bear,  are  external  forces. 

The  load  on  a  roof  truss  must  include,  besides  its  own  weight,  the 
weights  of  roof  covering  and  of  the  lintels.  In  addition  to  this,  an 
allowance  for  wind  and  snow  must  be  made.  Wind  pressure  is  as- 
sumed to  be  uniformly  distributed  and  is  reckoned  as  so  many 
pounds  to  the  square  foot  normal  to  the  rafters.  Knowing  the 
length  of  the  rafters  and  the  space  between  them,  the  load  due  to 
the  wind  which  each  truss  is  to  bear  is  known. 

Authorities  fairly  well  agree  that  no  roof  should  be  designed 
to  stand  a  total  pressure  of  less  than  40  pounds  per  square  foot. 

25.  The  reactions  on  a  structure  of  its  supporting  foundations 
are  external  forces,  but  they  are  distinguished  from  the  external 
forces  constituting  the  load  by  calling  them  "  supporting  forces." 

26.  For  the  equilibrium  of  a  structure,  the  external  forces  con- 
stituting the  load  must  equal  the  supporting  forces,  and  the  exter- 
nal and  internal  forces  must  balance. 

27.  In  the  graphic  method  of  determining  the  stresses  in  the 
members  of  the  framed  structures  here  to  be  considered,  the  loads 


FRAMED  STRUCTURES. — RECIPROCAL  DIAGRAM  363 

are  to  be  applied  at  the  joints.  If  the  true  point  of  application  of  a 
load  on  a  member  be  at  some  point  intermediate  between  its  joints, 
parallel  forces  equivalent  to  the  load  will  be  substituted  at  the 
joints.  If  the  point  of  application  of  a  load  be  at  the  middle  of  a 
member,  the  equivalent  parallel  forces  at  the  joints  will  each  be 
equal  to  one-half  the  load;  if  the  point  of  application  be  other 
than  at  the  middle,  then  the  equivalent  parallel  forces  at  the  joints 
may  be  determined  by  moments. 

28.  Since  the  joints  are  to  be  considered  frictionless,  or  perfectly 
smooth,  the  pressures  at  the  points  of  contact  between  the  mem- 
bers and  a  joint  act  normal  to  the  surface,  and  their  lines  of  action 
must,  therefore,  pass  through  the  center  of  the  joint.     The  total 
action  of  -the  joint  will  then  be  a  single  force — the  resultant  of 
these  pressures  acting  in  line  of  one  of  the  members.     This  result- 
ant action  in  the  member  exerts  an  equal  and  opposite  action  at  the 
joint  at  the  other  end,  thus  maintaining  the  equilibrium  of  the 
joint.     It  is  evident,  then,  that  a  member  in  tension  tends  to  pull 
the  joints  at  its  ends  toward  each  other,  and  one  in  compression 
tends  to  separate  them. 

29.  Reciprocal   Diagram. — Since  there  is  equilibrium   at  each 
joint  of  a  framed  structure,  each  joint  center  may  be  considered  the 
point  of  application  of  the  load  on  the  joint  and  of  the  stress  forces 
in  its  members,  all  acting  in  the  plane  of  the  frame.     It  follows, 
therefore,  that  a  closed  polygon  may  be  constructed  whose  sides 
represent  these  forces  in  magnitude  and  direction.     Such  a  polygon 
is  known  as  the  Reciprocal  Diagram  of  the  joint,  and  is  nothing 
more  nor  less  than  the  polygon  of  forces.     Since  the  stresses  in  the 
members  of  a  joint  are  determined  from  its  reciprocal  diagram, 
such  diagram  is  also  called  a  "  stress  diagram." 

30.  If  a  joint  of  a  structure  be  selected  at  which  but  two  mem- 
bers meet,  and  the  load  on  the  joint  be  known,  the  reciprocal  dia- 
gram of  the  joint  may  be  drawn  by  means  of  the  triangle  of  forces, 
and  the  magnitude  and  direction  of  the  stresses  in  the  members 
can  thus  be  determined.     With  this  knowledge  gained,  the  recip- 
rocal of  another  joint  may  be  drawn,  provided  that,  if  more  than 
two  members  meet  at  the  joint  there  must  be  known,  in  addition  to 
the  external  forces  acting  at  the  joint,  the  stresses  in  all  the  mem- 
bers except  two. 


364 


ELEMENTS  OF  GRAPHIC  STATICS 


31.  As  a  preliminary  to  the  construction  of  the  reciprocal  dia- 
grams of  the  joints  of  a  framed  structure,  a  scale  drawing  of  the 
structure  itself  must  be  carefully  made,  which  is  known  as  the 
Frame  Diagram. 

32.  The  complete  preparation  of  the  frame  diagram  comprises 
the  following: 

(1)  The  determination  of  the  magnitude  and  direction  of  the 
total  load  at  each  joint,  replacing  any  load  that  may  be  applied  be- 
tween two  joints  by  equivalent  parallel  forces  at  the  joints.  These 
equivalent  parallel  forces  are  determined  in  the  same  manner  as 


that  employed  in  determining  the  support  reactions  of  a  loaded 
simple  beam. 

(2)  The  determination  of  the  supporting  forces,  or  support  re- 
actions.    These  reactions  can  be,  and  often  are,  determined  from 
the  reciprocal  diagram,  but 'their  predetermination  affords  a  check 
as  to  accuracy,  and  not  infrequently  furnishes  the  known  external 
force  at  a  joint  where  but  two  members  meet,  thus  providing  a 
starting  point  for  the  construction  of  the  reciprocal  diagram. 

(3)  The  correct  lettering  of  the  diagram  according  to  the  Bow 
system. 

(4)  The  marking  with  arrowheads  of  all  the  external  forces,  giv- 
ing to  each  its  value,  and  taking  care  that  the  arrowheads  do  not 
cross  the  lines  of  the  frame  diagram. 

33.  It  should  be  noted  that  when  all  the  loads  and  support  reac- 
tions are  vertical,  loads  coming  over  the  supports  of  the  structure 


FRAMED  STRUCTURES. — RECIPROCAL  DIAGRAM  365 

are  omitted,  as  they  have  no  influence  on  the  stresses  in  the  mem- 
bers. Such  loads  must  be  deducted  from  the  total  support  reac- 
tions in  order  to  obtain  the  proper  values  of  R^  and  R2  to  be  used 
in  the  determination  of  the  stresses.  V 

34.  Figure  10  represents  a  "  King  Post  Truss  "  with  a  total  uni- 
form load  of  8  tons  on  one  side  and  4  tons  on  the  other.     The  loads 
are  applied  to  the  joints  as  follows : 

Of  the  4  tons  on  the  left  hand  rafter  we  may  assume  one-half  of 
it  borne  by  the  member  AF  and  the  other  half  by  the  member  BG. 
Distributing  these  two  loads  of  two  tons,  giving  a  half  of  each  to 
the  ends  of  its  member,  we  get  a  total  load  of  2  tons  at  joint  1, 
one  ton  at  the  apex  end  of  the  rafter,  and  one  ton  over  the  left 
support.  Proceeding  in  a  precisely  similar  manner  with  the  right 
hand  rafter,  we  get  a  total  of  3  tons  at  the  apex  of  the  rafters,  4 
tons  at  joint  3,  and  2  tons  over  the  right  support.  Rejecting  the 
loads  over  the  supports,  and  denoting  the  span  by  a,  we  have : 

E1X^  =  ^X^  +  3x|  +  4x|,  whence  R±  —  4  tons, 

and#2Xa  =  4X^  +  3x|  +  2x|,  whence  R2  —  5  tons. 

These  are  the  values  of  R±  and  R2  to  be  used  in  determining  the 
stresses  in  the  members,  but  the  total  wall  reactions  are  5  tons  and 
7  tons  for  R1  and  R2,  respectively. 

35.  Drawing  the  Reciprocal  Diagram. — In  drawing  the  recipro- 
cal diagram  we  proceed  as  follows : 

(1)  Select  a  force  scale — as  large  as  possible — and  draw  the 
force  diagram  of  the  external  forces,  marking  the  beginning  and 
ending  of  a  line  representing  a  force  with  the  small  letters  of  the 
alphabet  corresponding  to  the  capital  letters,  taken  in  clockwise 
order,  found  in  the  spaces  flanking  that  force  in  the  frame  dia- 
gram.    This  diagram  represents  the  magnitudes  and  directions  of 
the  external  loads  on  the  joints  and  of  the  support  reactions,  and, 
if  properly  drawn,  forms  a  closed  polygon.     In  the  most  frequent 
case,  that  of  vertical  loads,  the  polygon  resolves  itself  into  a  straight 
line,  vertical  in  direction,   as  already  explained.     Determine  the 
magnitude  of  the  support  reactions,  either  by  moments  or  by  the 
method  of  the  funicular  polygon. 

(2)  Choose  a  joint  at  which  but  two  members  meet,  and  resolve 
the  external  force  acting  at  the  joint  along  the  directions  of  the 


366 


ELEMENTS  OF  GRAPHIC  STATICS 


members,  lettering  the  components  clockwise  with  the  small  letters 
of  the  alphabet  corresponding  to  the  capital  letters  denoting  the 
members  in  the  frame  diagram.  This  will  give  a  triangle,  from 
which  the  magnitude  and  direction  of  the  stresses  in  the  two  mem- 
bers may  be  determined.  The  joint  at  one  of  the  supports  usually, 
but  not  always,  furnishes  a  starting  point,  the  known  reaction  at 
the  joint  being  the  external  force. 

(3)  One  of  the  stresses  thus  found  is  now  combined  with  the  ex- 
ternal or  other  known  force  at  the  next  joint  and  resolved  along  the 
directions  of  the  other  members  of  the  joint,  giving  a  polygon  of 


forces  from  which  the  stresses  in  two  more  members  may  be  found. 
This  process  is  continued  until  the  stresses  in  all  the  members  are 
determined. 

36.  It  should  be  noted  that  in  drawing  the  force  polygon  of  a 
joint  of  n  forces,  there  are  2n  conditions  to  be  satisfied,  viz. : 
n  magnitudes  and  n  directions,  and  unless  n  —  2  of  these  condi- 
tions are  known,  the  data  is  insufficient  to  construct  the  polygon. 

EXAMPLE  I. — Suppose  each  rafter  of  the  roof  truss  of  Fig.'  11  to 
support  an  evenly  distributed  load  of  6  tons ;  it  is  required  to  find 
the  magnitude  and  kinds  of  stresses  in  the  members  of  the  truss. 

We  commence  by  apportioning  the  loads  to  the  joints  of  the  frame 
diagram.  In  doing  so,  we  find  a  total  load  of  6  tons  at  the  ridge 
and  a  load  of  3  tons  directly  over  each  supporting  wall.  As  these 


FRAMED  STRUCTURES. — KECIPROCAL  DIAGRAM  367 

latter  do  not  affect  the  stresses  in  the  members,  they  will  be  omitted 
from  any  further  discussion,  remembering  that  if  the  total  reac- 
tions at  the  walls  are  required,  Rl  and  R2  must  each  be  increased 
by  3  tons. 

The  truss  being  symmetrical,  the  support  reactions,  R^  and  R2, 
are  the  same  and  each  equal  to  half  of  the  six  tons  at  the  ridge. 

Letter  the  frame  diagram,  as  shown,  ignoring  the  vertical  loads 
over  the  supports,  so  that  R2  will  be  known  as  BC,  and  R^  as  CA. 

Adopt  a  load  scale  of  J  inch  to  the  ton. 

Since  each  of  the  three  joints  of  the  frame  has  but  two  members 
and  a  known  external  force,  either  may  be  used  as  a  starting  point. 
Selecting  the  joint  at  the  left  support,  we  draw  ca,  equal  to  f  of 
an  inch,  to  represent  the  left  reaction,  CA,  and  we  measure  it  up- 
ward from  c  to  a  because  R^  acts  upward.  From  a  and  c  draw 
lines  parallel  to  the  members  AD  and  DC,  respectively.  They  in- 
tersect at  d,  giving  the  triangle  cad  as  the  triangle  of  forces,  or 
the  reciprocal  diagram,  for  the  joint  at  the  left  support.  Hence, 
ad  and  dc  represent  in  magnitude  and  direction  the  stresses  in  the 
members  AD  and  DC.  Knowing  the  direction  of  Rlf  as  repre- 
sented by  ca,  the  direction  of  the  stresses  in  the  two  members  are 
known  by  taking  the  sides  of  the  triangle  in  order,  as  shown  by 
the  arrows.  The  stress  in  AD  acts  in  the  direction  ad,  and  that  in 
DC  in  the  direction  dc.  Indicate  these  directions  by  placing  arrow- 
heads on  the  two  members  at  points  close  to  the  joint.  Knowing 
that  the  stress  in  one  end  of  a  member  is  opposed  by  an  equal  and 
opposite  stress  in  the  other  end,  we  may  now  place  arrowheads  at 
the  other  ends  of  the  members  AD  and  DC  accordingly,  as  shown. 

To  draw  the  reciprocal  of  the  joint  at  the  right  support,  we  draw 
be  upward,  and  make  it  f  of  an  inch  in  length,  to  represent  the 
right  reaction,  BC,  in  magnitude  and  direction.  From  b  and  c 
draw  lines  parallel  to  the  members  DB  and  CD,  respectively.  They 
intersect  at  d,  giving  the  triangle  bed  as  the  reciprocal  of  the  joint 
at  the  right  support.  Hence,  cd  and  db  represent  in  magnitude 
and  direction  the  stresses  in  the  members  CD  and  DB.  The  arrows 
within  the  triangle  bed  show  the  direction  of  these  stresses  and  they 
are  marked  with  arrowheads  accordingly  in  the  frame  diagram. 

By  scale  measurements  of  the  reciprocals,  the  stresses  in  the 
rafters  are  each  found  to  be  4.4  tons,  and  in  the  tie,  CD,  the  stress 
is  3.2  tons. 


368 


ELEMENTS  OF  GRAPHIC  STATICS 


The  magnitudes  and  directions  of  the  stresses  in  all  the  members 
are  now  known  and  we  have  yet  to  determine  only  their  kinds. 

37.  Kule  for  Determining  the  Kind  of  Stress  in  a  Member. — 
If,  after  determining  the  directions  of  the  stresses  in  all  the  mem- 
bers by  means  of  the  reciprocals  of  the  joints,  and  after  marking 
the  ends  of  the  members  with  arrowheads  accordingly,  it  is  found 
that  the  arrowheads  of  a  member  point  toward  its  joints,  the 
member  is  then  in  compression  and  is  a  strut;  if  they  point  away 
from  the  joint,  the  member  is  in  tension  and  is  a  tie. 

Hence,  it  is  found  that  the  rafters,  AD  and  BD,  are  in  com- 
pression,, and  the  member  CD  is  in  tension. 


38.  If,  to  the  frame  of  Fig.  11,  we  add  a  vertical  rod,  called  the 
<(  King  Post,"  we  obtain  the  frame  of  Fig.  12. 

Proceeding,  as  before,  we  determine  the  stress  in  AD  by  means 
of  the  triangle  cad. 

To  draw  the  reciprocal  of  the  joint  ABEDA  at  the  ridge,  we 
measure  ab  downward,  and  make  it  1.5  inches  long  to  represent, 
to  the  chosen  scale  of  0.25  inch  to  the  ton,  the  magnitude  and  direc- 
tion of  the  external  force  of  6  tons.  The  stress  in  the  rafter  AD 
has  been  found  to  be  4.4  tons,  so  we  draw  ad  parallel  to  AD,  and 
make  it  1.1  inches  long  to  represent  the  4.4  tons  to  scale.  From 
&  we  draw  a  line  parallel  to  BE  and  we  find  that  it  intersects  ad  at  d. 
This  shows  that  d  and  e  are  one  and  the  same  point,  and  that 
there  is  no  stress  in  the  vertical  post  DE.  Such  a  member  is  called 
"  redundant." 


FRAMED  STRUCTURES. — RECIPROCAL  DIAGRAM  369 

39.  As  a  further  illustration  we  will  consider  the  King-post  truss 
of  Fig.  10,  p.  364,  reproduced  in  Fig.  13.     The  loads  and  sup- 
port reactions  were  found  to  be  as  shown. 

Knowing  the  force  EA,  or  R19  we  commence  at  the  joint  at  the 
left  support  and,  with  a  scale  of  0.25  inch  to  the  ton,  obtain  eaf 
as  the  triangle  of  forces  for  the  joint.  Marking  with  arrowheads 
the  kind  of  stresses  in  the  members  AF  and  FE,  we  proceed  to  the 
next  joint  in  clockwise  order,  that  at  which  the  load  of  2  tons  is 
applied. 

Of  the  eight  conditions  of  this  joint,  the  four  directions  and 
two  of  the  magnitudes  are  known,  the  magnitude  of  FA  having  just 
been  determined.  We  then  readily  obtain  abgf  as  the  force  poly- 
gon of  the  joint.  Marking  with  arrowheads  the  kinds  of  stresses 
in  the  members  BG  and  BF,  we  proceed  in  clockwise  order  to  the 
other  joints,  obtaining  for  the  joint  at  the  ridge  the  force  polygon 
g~bch;  for  the  joint  HCDI  the  polygon  hcdi',  and  for  the  joint  at 
the  left  support  the  triangle  ide,  thus  completing  the  determination 
of  the  stresses  in  all  the  members. 

It  is  observed  that  each  of  the  force  polygons  of  the  joints, 
when  taken  in  order,  contains  a  side  of  the  one  immediately  pre- 
ceding it ;  hence,  each  polygon  can  be  built  upon  the  one  preceding 
it,  and  thus  produce  one  figure  which  will  contain  all  the  sides, 
and  be  a  graphic  representation  of  the  magnitudes  and  directions 
of  all  the  external  forces  and  internal  stresses  of  the  structure. 
Such  a  figure  is  known  as  the  Reciprocal  Diagram  of  the  structure. 

The  reciprocal  diagram  of  the  truss  of  Fig.  13  is  shown  in  Fig. 
14,  the  load  line  ad  having  first  been  drawn  and  the  lines  repre- 
senting the  stresses  then  drawn  in  their  regular  order.  The  mag- 
nitudes of  the  stresses  were  measured  and  found  to  be  as  shown  in 
the  table. 

40.  It  is  convenient  sometimes  to  determine  the  stress  in  a  mem- 
ber by  the  "  Ritter  Section  "  method.     This  method  depends  upon 
the  principle  that,  at  any  imaginary  section  of  a  frame  there  is 
equilibrium  between  the  external  forces  on  one  side  of  the  section 
and  the  stress  forces  on  the  same  side  that  are  in  the  members  cut 
by  the  section,  and  that,  therefore,  the  algebraic  sum  of  the  mo- 
ments of  these  forces  about  any  point  in  the  plane  of  the  frame 
is  zero. 

In  Fig.  13,  suppose  an  imaginary  section  xy  to  be  taken.     This 
24 


Fiy. 


FRAMED  STRUCTURES. — KECIPROCAL  DIAGRAM 


371 


section  cuts  the  members  EF,  BG,  and  GF.  Then,  the  left  support 
reaction  and  the  force  of  2  tons  are  balanced  by  the  stresses  in  the 
members  EF,  BG,  and  GF.  Taking  moments  about  the  left  sup- 
port, we  have :  GF  X  r  =  2  X  s,  whence  Stress  in  GF  =  ^ .  The 
frame  diagram  of  Fig.  13  was  constructed  to  a  scale  of  -^  inch  to 
the  foot,  and  since  s  and  r  measure  6.25  and  7.5  feet,  respectively, 

X6.25 


we  get  Stress  in  GF  = 


7.5 


=  1.67  tons. 


41.  Any  attempt  to  put  arrowheads  on  the  reciprocal  diagram  to 
indicate  the  kinds  of  stresses  in  the  members  results  in  nothing  but 
confusion.  If  the  kind  of  stress  in  a  member  cannot  be  discovered 
by  the  eye,  the  polygon  of  forces  for  the  joint  in  question  must  be 
drawn  as  was  done  in  connection  with  Fig.  13. 


.15. 


42.  The  stresses  in  the  members  of  the  braced  cantilever  of 
Fig.  15,  having  a  concentrated  load  of  2  tons  at  its  outer  extremity, 
may  be  found  as  follows : 

To  a  scale  of  -fa  inch  =  -fa  ton,  lay  off  ab  vertical  and  J  inch 
in  length  to  represent  the  load  of  2  tons.  From  b  draw  a  parallel 
to  BC  and  from  a  a  parallel  to  CA,  intersecting  at  c.  Then,  abc 
is  the  triangle  of  forces  for  the  equilibrium  of  the  joint  ABC.  The 
force  AB,  acting  downward,  is  denoted  in  direction  and  magnitude 
by  ab  in  the  triangle,  and  the  stresses  in  the  other  members,  taken 
in  order,  act  in  the  directions  be  and  ca,  and  their  magnitudes  are, 
of  course,  denoted  by  the  lengths  of  these  lines  to  the  chosen  scale. 


372  ELEMENTS  OF  GRAPHIC  STATICS 

Mark  these  directions  by  arrowheads  at  the  joint  ABC.  The  di- 
rections of  the  stress  in  the  upper  end  of  BC  and  in  the  outer  end 
of  CA  being  now  known,  it  follows  that  the  stresses  in  the  other 
ends  of  these  members  are  equal  and  opposite,  and  arrowheads  must 
be  at  once  placed  to  indicate  them.  We  now  have  sufficient  data 
to  draw  the  force  diagram  for  the  joint  CBD.  The  stress  in  the 
lower  end  of  CB  has  just  been  marked  to  act  in  the  direction  cb. 
From  b  and  c  draw  parallels  to  BD  and  DC,  respectively,  intersect- 
ing at  d;  then  the  triangle  cbd  is  the  force  diagram  for  the  joint 
CBD,  and  bd  and  dc  are  the  directions  of  the  stresses  in  the  mem- 
bers BD  and  DC,  and  have  been  marked  accordingly  in  the  frame 
diagram.  Arrowheads  are  at  once  placed  at  the  other  ends  of  BD 
and  DC  to  denote  the  directions  of  the  equal  and  opposite  stresses 
at  the  wall  and  at  the  joint  ACDE,  respectively.  The  stresses  in 
AC  and  CD  at  the  joint  ACDE  are  now  known,  and  their  direc- 
tions are  ac  and  cd,  respectively.  From  d  draw  a  parallel  to  DE 
and  from  a  a  parallel  to  EA.  They  intersect  at  e,  and  de  and  ea 
are  the  directions  and  magnitudes  of  the  stresses  in  the  members 
DE  and  EA,  respectively.  It  should  be  noted  that,  in  the  recip- 
rocal diagram,  the  stress  line  of  one  member  may  lie  wholly  or 
partly  on  the  stress  line  of  another  member,  as  was  here  instanced  in 
the  stresses  of  AC  and  EA.  The  stresses  in  the  different  members 
were  measured  to  scale  on  the  reciprocal  diagram  and  marked  on 
the  members  of  the  frame. 

The  horizontal  outward  pull  of  2.95  tons  at  the  upper  joint  at 
the  wall  occasions  an  equal  and  opposite  reaction.  At  the  lower 
wall-joint  there  is  a  horizontal  thrust  of  2  tons  and  a  diagonal  thrust 
of  2.25  tons.  The  horizontal  component  of  this  diagonal  thrust  is 
hd  =  ac  =  0.95  tons,  making  the  two  reactions  equal  but  opposite 
in  direction. 

43.  the  braced  cantilever  of  Fig.  16  is  25  feet  long,  10  feet  deep, 
and  uniformly  loaded  on  the  top  with  100  pounds  per  foot  run. 

In  apportioning  the  total  load  of  2500  Ibs.,  it  will  be  observed 
that  the  member  BJ,  being  but  half  the  length  of  the  members 
CH  and  DF,  sustains  but  J  of  the  total  load,  CH  and  DF  each  sus- 
taining f . 

The  apportionment  of  the  load  will  therefore  be :  500  pounds  at 
the  joint  at  the  outer  end,  1000  pounds  at  the  joint  CDFGH,  750 
pounds  at  the  joint  BCPIIJ,  and  250  pounds  at  the  joint  ABJ  at 


FRAMED  STRUCTURES. — EECIPROCAL  DIAGRAM 


373 


the  wall.     This  latter  load,  having  no  influence  on  the  stresses  in 
the  members,  is  rejected. 

Commencing  at  the  joint  DEF,  and  with  a  scale  -£$  inch  =  100 
pounds,  we  get  def  as  the  force  diagram  for  the  equilibrium  of  the 
joint,  and  at  once  mark  arrowheads  on  the  frame  indicating  the 
directions  of  the  stresses  in  EF,  FD,  FE,  and  DF.  Having  now 
the  magnitude  of  the  stress  FE  at  the  joint  FEG,  we  obtain  feg  as 
the  force  diagram  of  the  joint.  Marking  with  arrowheads  on  the 
frame  the  direction  of  the  stresses  thus  obtained,  we  find  that  at  the 


I. 


joint  CDFGH  the  force  CD  and  the  stresses  DF  and  FG  are  known 
and,  therefore,  the  force  polygon  cdfgh  is  readily  obtained.  The 
reciprocal  diagram  may  now  be  completed  without  difficulty. 

44.  The  Warren  Girder  of  Fig.  17  has  a  span  of  27  feet,  and 
loaded  with  1-J  tons  per  foot,  making  a  total  load  of  36  tons.  The 
members  AF  and  DL  each  sustain  but  J  of  the  load,  and  in  the 
apportionment,  y1^,  or  3  tons,  fall  over  the  support  at  each  end,  and 
the  remainder  of  the  load  as  shown. 

At  each  end  of  the  top  flange  there  is  equilibrium  under  the 
action  of  the  external  force  of  3  tons  in  line  with  the  upright  mem- 
ber and  the  stresses  in  the  two  members  at  right  angles  to  each 


374 


ELEMENTS  or  GEAPHIC  STATICS 


other.  Evidently  the  stress  in  each  of  the  uprights  is  3  tons,  there- 
fore there  cannot  be  any  stress  in  the  members  AF  and  DL  if  the 
equilibrium  is  to  be  maintained.  To  indicate  that  there  is  no  stress 
in  AF,  the  letters  a  and  /  must  be  placed  at  the  same  point  in  the 
reciprocal  diagram,  and  because  of  the  absence  of  stress  in  DL,  the 
letters  d  and  I  are  at  the  same  point.  It  should  not  be  forgotten 
that  the  two  external  forces  of  3  tons  at  the  ends,  coming  directly 
over  the  supports,  are  neglected,  and  that  the  reactions  of  15  tons 


i2.\+ons 


9  \tons    3\+or# 


\tons    3\ 

f    £>   1 


€.2  tons. 
Ci/-  / 2.5  tons. 


Fiy.  /. 


each  are  due  solely  to  the  loads  of  9  tons,  12  tons,  and  9  tons,  the 
forces  which  occasion  the  stresses  in  the  members.  In  the  con- 
struction of  the  Warren  Girder  the  end  uprights  and  the  members 
AF  and  DL  are  omitted. 

To  the  scale  of  -J  inch  =  3  tons,  set  off  ef  to  represent  in  mag- 
nitude and  direction  the  left  reaction,  EF,  of  15  tons.  From  / 
and  e  draw  parallels  to  FG  and  GE,  respectively,  intersecting  at  g. 
efg  is  then  the  triangle  of  forces  for  the  joint  at  the  left  support. 
For  the  joint  GFABH  we  get  gfabJi  as  the  polygon  of  forces,  and 
so  on  to  the  completion  of  the  reciprocal  diagram. 

The  triangles  of  the  Warren  Girder  being  equilateral,  the  stresses 
found  from  the  reciprocal  diagram  may  easily  be  checked.  Thus, 
the  stress  in  KL  —  15  sec  30°  =  17.32  tons. 


FRAMED  STRUCTURES. — EECIPROCAL  DIAGRAM 


375 


45.  The  Linville,  or  N,  Girder,  of  Fig.  18  is  irregularly  loaded 
on  the  bottom  flange  as  shown. 

To  a  scale  of  ^  inch  =  1  ton,  set  off  ab  and  be  to  represent  in 
magnitude  and  direction  the  loads  of  20  tons  and  30  tons,  respec- 
tively. 

For  convenience  the  frame  is  lettered,  in  this  instance,  in  contra- 
clockwise  order,  and  the  forces  at  the  joints  will  be  considered  like- 
wise. 


OK *JB*  31.5 ton*; 


Selecting  a  pole,  o,  we  obtain,  by  means  of  the  funicular  polygon, 
cd  and  da  for  the  reactions  R2  and  Rlt  respectively. 

For  the  reason  already  given,  there  can  be  no  stress  in  the 
members  CL  and  EA,  and  this  is  indicated  by  placing  Z,  in  the 
reciprocal  diagram,  at  the  same  point  with  c,  and  e  at  the  same 
point  with  a. 

At  the  joint  at  the  middle  of  the  top  flange  there  is  a  state  of 
equilibrium  under  the  action  of  the  stresses  in  the  members  ID, 
DH,  and  HI,  HI  being  at  right  angles  to  each  of  the  others.  There 
cannot,  therefore,  be  any  stress  in  HI,  and  h  and  i  will  fall  at  the 
same  point  in  the  reciprocal  diagram. 


376 


ELEMENTS  OF  GRAPHIC  STATICS 


The  members  CL  and  EA  add  rigidity  to  the  frame,  and  HI 
resists  the  tendency  of  the  top  flange  to  bend. 

For  the  equilibrium  of  the  joint  LDK  we  get  the  triangle  Idle. 
For  the  joint  CLKJB  we  get  the  force  polygon  elk  ft,  and  so  on  to 
the  completion  of  the  reciprocal  diagram. 

46.  The  Fink  Truss  of  four  panels,  Fig.  .19,  has  a  span  of  64 
feet,  a  depth  of  12  feet,  and  is  uniformly  loaded  with  1.5  tons  per 
foot,  making  96  tons  in  all.  The  figure  is  constructed  to  a  scale  of 
^5-  of  an  inch  to  the  foot,  and  the  load  scale  is  taken  as  -£$"  =  1  ton. 


tons 


The  apportionment  of  the  load  places  12  tons  over  each  support, 
and  they  are  rejected. 

There  is  insufficient  data  to  begin  at  the  joint  at  the  left  support, 
but  by  means  of  the  Eitter  section,  xy,  and  moments  about  the  left 
support,  we  obtain: 

Stress  in  EH  X  32  sin  6  =  24  X  16,  whence,  Stress  in  EH  =  20 
tons,  and  it  is  evident  that  the  members  EG,  EM,  and  EN  are  sub- 
jected to  the  same  stress. 

On  the  load  line  set  off  ab,  be,  and  cd,  each  equal  to  ||  inch  to 
represent  the  external  loads  of  24  tons.  Then,  de  =  ea  =  the  re- 
actions to  scale. 

From  e  draw  eg  parallel  to  EG,  and  make  it  f  $  inch  long  to  rep- 
resent the  stress  of  20  tons.  Commencing  at  e,  we  now  get  eafge 


FRAMED  STRUCTURES. — KECIPROCAL  DIAGRAM  377 

as  the  stress  polygon  for  the  joint  at  the  left  support.  Knowing  af, 
the  stress  polygon  abif  for  the  joint  ABIF  is  readily  drawn,  and 
so  on  to  the  completion  of  the  reciprocal  diagram. 

By  means  of  the  Eitter  section  ut  and  moments  about  the  upper 
middle  joint,  we  obtain: 

Stress  in  JE  X  32  sin  a  +  24  X  16  =  36  X  32, 

TI,        (36  X  32  —  24  X  16)^/73 
whence,  Stress  in  JE  =  -  Q6  =68.35  tons, 

which  agrees  with  the  scale  length  of  je  of  the  reciprocal  diagram. 

The  stresses  in  AF,  BI,  CL,  and  DO  are  shown  by  the  diagram  to 
be  80  tons  each.  This  may  be  checked  by  the  Ritter  sections,  ut 
and  vz. 

Thus,  for  the  section  ut  and  moments  about  the  joint  GI1E,  we 
obtain:  Stress  in  BI  X  12  —  36  X  16  +  Stress  in  JE  X  6  cos  a, 

whence  Stress  in  BI  —  48  +  68.35  X  -4y  =  80  tons. 

From  the  section  vz  and  moments  about  the  joint  FIHG,  we 
have: 
Stress  in  EG  X  2\/73  sin  (6  —  a)  +  Stress  in  AF  X  6  =  36  X  16, 

whence,  Stress  in  AF  =  -          —  =  80  tons. 

47.  The  reciprocal  diagrams  of  all  the  frames  that  have  been  con- 
sidered have  closed,  and  the  frames,  therefore,  complete.     It  will 
be  noticed  that,  in  the  cantilever  frames,  the  number  of  members 
equals  twice  the  number  of  joints  minus  four,  and  that  in  the 
frames  supported  at  the  ends  the  number  of  members  equals  twice 
the  number  of  joints  minus  three. 

48.  A  frame  having  more  than  the  requisite  number  of  members 
is  redundant;  that  is,  it  contains  more  than  the  necessary  number 
of  members  to  prevent  distortion. 

49.  A  deficient  frame  is  one  that  has  not  sufficient  members  to 
prevent  distortion. 

PROBLEMS. 

4.  Draw  the  reciprocal  diagram  of  the  Warren  Girder,  Fig.  20, 
the  triangles  being  equilateral.  Span,  42  ft.,  and  a  uniformly  dis- 
tributed load  of  1.5  tons  per  foot.  Tabulate  the  stresses,  and  indi- 


378 


ELEMENTS  OF  GRAPHIC  STATICS 


cate  their  kinds  in  the  frame  diagram.     Scales :  load,  TV  =  1  ton; 
linear,  TV  =  1'. 


5.  Draw  the  reciprocal  diagram  of  the  roof  truss  of  Fig.  21, 
which  is  loaded  as  shown.  Tabulate  the  stresses,  and  indicate  their 
kinds  in  the  frame  diagram.  Scales:  Load,  \  inch  =  1  ton;  lin- 
ear, Y1^  inch  =  1  foot. 


6.  The  Fink  truss  of  Fig.  22  has  a  span  of  56  ft.,  a  depth  of  12 
feet,  and  is  uniformly  loaded  with  1.5  tons  per  foot.  Draw  the  re- 
ciprocal diagram  to  the  scale  of  Ty  =  1  ton.  Tabulate  the 
stresses,  and  indicate  their  kinds  in  the  frame  diagram.  Linear 
scale,  TV  inch  =  1  foot. 


I 


7.  The  braced  cantilever  of  Fig.  23  is  20  ft.  long,  9  ft.  deep,  and 
uniformly  loaded  on  the  top  with  120  Ibs.  per  foot.  Draw  the 
reciprocal  diagram  to  a  scale  of  -^"  —  100  Ibs.  Tabulate  the 
stresses,  and  indicate  their  kinds  in  the  frame  diagram.  Show  that 


FRAMED  STRUCTURES. — RECIPROCAL  DIAGRAM 


379 


the  reactions  at  the  wall  are  equal,  but  opposite  in  direction.     Lin- 
ear scale,  T3-g-  inch  =  1  foot. 


8.  Draw  the  reciprocal  diagram  of  the  braced  cantilever  of  Fig. 
24,  tabulating  the  stresses,  and  indicating  their  kinds  in  the  frame 


diagram.  Give,  in  magnitude  and  direction,  the  resultant  stress 
on  the  pin  of  the  upper  joint  at  the  wall.  Scales :  Load,  TV  inch 
=  80  Ibs. ;  linear,  J  inch  =  1  foot. 


APPENDIX 


APPENDIX 


MASS  AND  MEASUREMENT  OF  FORCE. 

The  Mass  of  a  body  is  the  invariable  quantity  of  matter  it  contains. 

The  Weight  of  a  body  is  the  name  given  to  the  pressure  which  the 
attraction  of  the  earth  causes  a  body  to  exert;  or,  in  other  words,  it  is 
the  force  with  which  the  earth  attracts  a  body. 

In  the  measurement  of  force  there  are  two  recognized  units,  viz.,  the 
Absolute  Unit  and  the  Gravitation  Unit. 

Definition. — The  Absolute  Unit  of  Force,  called  in  England  the  Poundal, 
is  that  force  which,  acting  for  one  second  on  a  mass  of  one  pound,  im- 
parts to  it  a  velocity  of  one  foot  per  second. 

Definition. — The  Gravitation  Unit  of  Force,  called  the  pound,  is  the 
force  required  to  support  a  mass  of  one  pound  against  the  attractive 
force  of  gravity. 

It  has  been  found  by  experiment  that  if  a  body  weighing  one  pound 
be  allowed  to  fall  freely  for  one  second  it  will  acquire  a  velocity  of 
about  32.2  feet  per  second,  this  32.2  feet  per  second  being  the  accelera- 
ting force  of  gravity,  universally  denoted  by  g.  It  follows  from  the 
definitions  that  the  gravitation  unit  of  force  is  g  times  as  great  as  the 
absolute  unit;  that  is,  one  pound  is  equal  to  g  poundals. 

The  mass  of  a  body  being  an  invariable  quantity,  the  absolute  unit  is 
therefore  invariable,  and  is  the  unit  employed  in  theoretical  mechanics, 
in  the  solution  of  problems  in  which  the  results  are  to  show  a  high 
degree  of  accuracy  and  are  to  be  independent  of  locality. 

Since  the  accelerating  force  of  gravity  is  a  variable  quantity,  increas- 
ing slightly  as  we  pass  from  the  equator  toward  the  poles,  the  gravi- 
tation unit  of  force  is  not  constant.  The  variation  is  scarcely  appre- 
ciable, but  it  is  none  the  less  true  that  the  gravitation  unit  does  not 
possess  the  quality  of  invariability.  Notwithstanding  this  fact,  the 
gravitation  unit  of  force  is  the  one  adopted  by  engineers  in  the  science 
of  practical  mechanics,  in  which  forces  are  measured  by  the  weights 
they  will  support,  the  unit  of  measurement  being  a  pound. 

The  pound  weight  has  long  been  the  standard  of  reference  in  the 
measurement  of  force.  The  pressure  of  steam  in  a  boiler  is  reckoned 
in  pounds  per  square  inch;  the  tension  in  a  string,  the  power  of  the  blow 
of  a  steam  hammer,  the  force  necessary  to  draw  a  railway  train,  are  all 
measured  in  pounds,  and  therefore  the  gravitation  unit  is  the  convenient 
one  for  use,  and  it  would  be1  very  inconvenient  to  use  the  absolute  unit 
in  calculations  for  practical  purposes,  though  its  use  in  theoretical  in- 
vestigations is  desirable. 


384  APPENDIX 

In  gravitation  measure,  the  unit  of  mass  is  the  quantity  of  matter  in 
a  body  weighing  g  pounds,  so  that  a  body  of  mass  1  weighs  g  pounds, 
and  a  body  of  mass  M  weighs  Mg  pounds.  Then,  if  M  denotes  the  mass 

W 
of  a  body  weighing  W  pounds,  we  shall  have,  W  =  Mg,  whence  M  =  —  ; 

9 

so  that  the  numerical  estimate  of  the  mass  of  a  body  is  derived  from  its 
weight,  and  although  the  weight  of  a  body  varies  as  it  is  carried  from 

W 
place  to  place,  the  ratio  —  remains  constant.     For,  let  W  be  the  weight 

9 

of  a  body  in  a  certain  locality  where  the  acceleration  of  gravity  is  g, 
and  let  W  be  the  weight  of  the  same  body  at  another  locality  where  the 

acceleration  of  gravity  is  g'.    We  would  then  have:  -— = —^  =  ^ , 

whence  -2"  —^-.     Therefore,  though  W  and  g  vary  with  the  locality, 
y        y 

TTT 

the  ratio  _   remains  constant,  and  the  numerical  value  of  M  for  the 
9 

same  body  is  constant  at  all  places. 

A  body  of  mass  M  moving  with  a  velocity  v  is  said  to  possess  momen- 
tum, or  quantity  of  motion,  which  is  measured  by  the  product  Mv. 

By  the  First  Law  of  Motion,  the  change  of  momentum  of  a  moving 
body  is  caused  by  the  action  of  a  force,  and,  by  the  Second  Law,  the 
rate  of  change  of  momentum  is  proportional  to  the  acting  force. 

Thus,  if  a  force  F  act  on  a  body  of  mass  M  for  a  period  of  t  seconds, 
changing  its  velocity  from  v±  to  v2,  then  M(v2  —  vt)  will  express  the 

change  in  momentum,  and  "*  2  ~  V{  will  be  the  measure  of  the  rate 
of  change  of  momentum.  Then,  F  =  - —  ,~  •  —  cMf,  in  which  f 
—  v-i  ~  vi  __  tne  acceleration  in  feet  per  second,  and  c  some  constant 

which  it  is  desirable  to  make  unity.  If  the  absolute  unit  of  force  be 
adopted,  then  M  =  l  and  f  =  l,  so  that  c  =  l,  and  we  have  F  =  Mf; 
that  is, 

Force  —  Mass  x  Acceleration. 

In  the  equation,  F  —  Mf,  the  force  F  is  expressed  in  poundals,  or  in 
absolute  measure.  It  is,  as  has  been  shown,  preferable  to  express 
forces  by  their  static  measure,  that  is,  by  the  weights  they  will  sup- 
port, so  in  order  that  the  fundamental  equation  shall  hold  for  gravita- 
tion units  it  must  be  modified  to  suit  them. 

Let  F  and  F'  be  the  static  measures  of  two  forces  that  produce  accel- 
erations f  and  f  on  a  mass  M.  Then,  by  the  Second  Law  of  Motion, 
F  :  F'  =  f  :  f.  If  F'  be  the  force  due  to  gravity,  viz.,  the  weight,  W,  of 
the  body,  then  the  acceleration  is  g,  and  we  shall  have  F  :  W=f  :  g, 

W 

whence  F  =  — f,  in  which  F  is  expressed  in  the  same  units  as  W,  that 
9 

W 
is,  in  the  gravitation  units  of  pounds,  the  invariable  ratio  —  denoting 

9 
the  mass. 


APPENDIX 


385 


CENTRIPETAL  AND  CENTRIFUGAL  FORCES. 

If  a  body  describes  a  circle  of  radius  r,  with  a  uniform  velocity  v,  the 
body  is  acted  on  by  a  force  tending  to  the  center  of  the  circle  the  acceler- 
ation of  which  is  — . 
r 

Let  PQ,  Pig.  1,  be  an  arc  of  the  circle  such  that  P  and  Q  are  very  near 
together.  Join  P  and  Q  with  the  center  of  the  circle:  then  the  angle 
POQ  is  very  small  and  will  be  denoted  by  a.  Let  t  denote  the  small 
interval  of  time  during  which  the  arc  PQ  was  described.  In  the  time 
t  a  velocity  parallel  to  PO,  and  equal  to  v  sin  «,  has  been  generated. 
The  acceleration,  f,  parallel  to  PO  is, 

_  v  sin  a  _  va  _  v    arc  PQ 

~          t        ~T  ~~  ~t'        r 

But  arc  PQ  —  space  described  in  time  t  =  vt,  hence  f  =  -  - . 


.  I. 


Then,  since  F  —  Mf,  we  have,  as  the  force  tending  to  the  center, 


r  gr 

That  is,  if  a  body  of  mass  M  describes  a  circle  of  radius  r,  with  uni- 
form velocity  v,  then,  whatever  the  forces  acting  on  the  body,  their 

Mv* 


resultant  tends  to  the  center  of  the  circle,  and  is  equal  to 


This 


inward  pull  on  the  body  toward  the  center  of  rotation  is  known  as 
centripetal  force.  The  body  being  attached  to  the  center,  this  inward 
pull  will  be  felt  at  the  center,  and  will  cause  a  reaction  on  the  center 
itself.  This  reaction  is  known  as  centrifugal  force.  Hence,  the  centrip- 
etal and  centrifugal  forces  acting  on  a  revolving  body  are  equal  and 


in  which  v  is  the 


opposite,  and  are  numerically  equal  to  _^-  — _ 

linear  velocity  of  the  body. 

If  the  angular  velocity  of  a  revolving  body,  that   is,   the   circular 
measure  of  the  angle  described  by  any  line  of  the  body  in  a  given  time, 

25 


386  APPENDIX 

be  given  and  be  denoted  by  «,  then  the  arc  described  in  the  given  time 
by  a  point  distant  r  from  the  center  will  be  rw.  But  the  length  of  the 
arc  is  the  linear  velocity  of  the  point  in  the  given  time;  hence,  the  equa- 
tion connecting  linear  and  angular  velocities  is  v  =  ru. 

Then,    Centripetal    Force  —  Centrifugal    Force  —  Wr*u*  —  Wa)*r  ,  in 

gr  g 

which  a)  is  the  angular  velocity  of  the  body. 


ACCUMULATED  WORK. 

A  moving  body  is  said  to  have  work  accumulated  in  it,  and  it  can  be 
made  to  do  work  by  parting  with  its  velocity. 

Let  a  body  of  mass  M  be  moving  with  a  velocity  v;  let  a  constant 
force  F,  acting  on  the  body  through  a  space  s,  bring  it  to  rest;  then  we 
shall  take  Fs  as  the  measure  of  the  work  accumulated  in  the  body. 

We  have  v2  =  2fs,  and  f  =  ~t  whence  v2=  ^.     Then, 

M  M. 

Accumulated  Work  —  Fs  =  ^ . 

& 

The  expression  _JL  is  called  the  Vis  Viva,  or  Kinetic  Energy,  of  the 

0 

mass  M,  and  is  the  exact  equivalent  of  the  work  accumulated  in  the 
body,  and  is  also  the  measure  of  the  work  the  force  F  will  have  to  do 
before  bringing  it  to  rest. 

Let  Ti  be  the  height  through  which  the  body  must  fall  to  acquire  the 
velocity  v.  Then,  since  W  =  Mg  and  vz  =  2gh,  we  have  for  the  accu- 
mulated work  in  the  body,  ^2 -_  _^2  _  _^  .  2gh  =  Wh. 

&  ag          tig 

Hence  we  say,  that  the  work  accumulated  in  a  moving  body  is  meas- 
ured by  the  product  of  the  weight  of  the  body  into  the  height  through 
which  it  must  fall  to  acquire  the  velocity. 


STEAM    DISTRIBUTION    BY   THE  SLIDE   VALVE 

AND 
THE    FORMATION    OF   THE    INDICATOR    DIAGRAM. 

In  figures  2  to  9  are  shown  the  important  positions  of  a  plain  slide 
valve  during  a  stroke  of  the  piston,  and  a  skeleton  arrangement  of  the 
piston-crank-eccentric-valve  mechanism.  The  valve  V  moves  to  and  fro, 
admitting  steam  first  to  one  side  of  the  piston  and  then  to  the  other 
from  the  steam  chest  8  through  the  steam  ports  s  and  s',  and  allowing 
the  used  steam  to  escape  through  the  same  ports  to  the  exhaust  port  e. 
The  varying  pressures  of  the  steam  thus  introduced  move  the  piston  P 
alternately  from  one  end  of  the  cylinder  to  the  other.  This  motion  is 
transferred  to  the  outside  of  the  cylinder  by  the  piston  rod  PH.  The 
connecting-rod  HC  connects  the  outside  end  of  the  piston  rod  to  the 
crank  pin  G  of  the  crank  OG.  As  G  is  constrained  to  move  about  0  as 


APPENDIX 


387 


a  center,   the  reciprocating  motion   of  the  piston   is   transmuted    into 
rotary  motion  of  the  crank. 

To  have  the  engine  turn  in  the  direction  indicated  by  the  arrow,  the 
eccentric  must  be  placed  ahead  of  the  crank  by  90°  plus  the  angle  of 
advance,  as  at  E.  In  the  actual  engine,  the  eccentric  and  crank  are 
each  attached  to  the  same  shaft,  but  in  order  to  show  the  mechanism 
more  clearly,  the  crank  pin  and  eccentric  circles  have  each  been  tufned 
through  90°  to  bring  them  in  the  plane  of  the  piston  and  valve  rods. 
The  movement  of  the  eccentric  is  transferred  to  the  valve  through  the 
eccentric  rod  EB  and  the  valve  rod  BV.  Thus  the  motion  of  the  piston 


Exhaust  Lead 


Fig.  2. 


to  make  that  motion 


causes  the  valve  to  so  distribute  the  steam 
continuous. 

The  dimensions  of  the  valve  may  easily  be  found  when  the  laps  are 
given.  Thus  in  Fig.  6,  by  laying  off  from  the  edge  of  the  steam  port 
the  steam  lap  on  the  steam  edge  of  the  valve,  which  in  this  case  is  the 
outside  edge,  and  the  exhaust  lap  on  the  exhaust  edge,  the  size  of  the 
valve  is  determined.  The  throw  of  the  eccentric  and  the  angular  ad- 
vance will  then  determine  how  much  lead  and  overtravel  the  valve  will 
have,  and  where  the  points  of  cut-off,  release,  compression,  and  admis- 
sion will  occur. 

Below  each  of  these  figures  are  the  horizontal  lines  ov  and  oV.  The 
distance  of  pb  and  p'b'  above  these  lines  represents  to  some  scale  the 
boiler  pressure.  Also,  the  lines  aa  and  a'a'  show  by  their  distances 


388 


APPENDIX 


above  ov  and  o'v'  respectively,  the  pressures  against  which  the  steam 
is  exhausted,  which,  in  this  case,  is  the  atmospheric  pressure. 

Any  point  on  this  diagram  will  represent  by  its  horizontal  distance 
from  the  end  of  the  diagram  the  position  of  the  piston,  and  from  the 
line  op  or  o'p'  the  volume  of  steam  in  the  cylinder.  The  vertical  dis- 
tances of  any  point  above  ov  and  aa,  or  above  o'v'  and  a'a',  will  repre- 
sent respectively  the  absolute  and  gauge  pressures  per  square  inch 
acting  on  the  piston.  The  upper  diagram  represents  what  is  taking 
place  on  the  left-hand  side  of  the  piston,  while  the  lower  diagram  is  a 
representation  of  what  is  taking  place  on  the  right-hand  side. 


Maximum  Opening  to  Steam 


b'-  —i/o'     Full  Opening  to  Exhousf 


a' 


Fig.  3. 


In  Fig.  2,  the  piston  is  at  the  head  end  of  the  stroke  with  the  valve 
open  to  steam  lead  at  s,  letting  the  steam  in  to  start  the  piston  on  its 
stroke;  while  at  s'  is  shown  the  exhaust  lead,  letting  the  steam  out  so 
as  not  to  retard  the  piston's  motion.  The  piston  being  in  its  initial  posi- 
tion and  about  to  start  on  the  forward  stroke,  the  height  of  the  point 
1  above  ov  indicates  the  pressure  which  urges  the  piston  forward,  and 
the  height  of  the  point  1'  above  o'v'  shows  the  pressure  tending  to  pre- 
vent the  forward  movement. 

The  valve  moves  toward  the  right  until  the  eccentric  radius  comes 
in  line  with  the  eccentric  rod,  the  piston  reaching  the  position  shown  in 
Fig.  3.  The  valve  now  has  the  maximum  opening  to  steam  at  s,  and 
full  opening  to  exhaust  at  s'.  The  ports  are  designed  to  let  the  steam 
out,  and  therefore  the  exhaust  should  be  fully  open  for  at  least  part  of 


APPENDIX 


389 


v'- 


Exhaust  Taking  Place 


Fig.  4. 


Is I 


Expansion  Taking  Place 


P'        Point  of  Compression 


Fig.  5. 


390 


APPENDIX 


a 

V  - 


Expansion  Taking  Place 


' p>  Compression  Taking  Place 


o' 


Fig.  6. 


6' 

6' 

L    ,  ,..> 

a'  a' 


v 


Point  of  Release 


Compression  Taking  Place 


Fig.  7. 


APPENDIX  391 

the  stroke.  This  is  accomplished  by  giving  the  valve  overtravel,  which 
is  the  distance  the  exhaust  edge  of  the  valve  goes  beyond  the  steam 
port  when  at  the  end  of  its  travel.  The  positions  of  the  points  2  and  2' 
relative  to  ov  and  o'v',  respectively,  show  the  pressures  on  the  piston. 

As  the  crank  continues  to  revolve,  the  valve  moves  toward  the  left. 
When  the  steam  edge  comes  in  line  with  the  steam  port,  Fig.  4,  cut-off 
occurs,  and  the  position  of  the  point  3  represents  the  pressure  at  that 
point  of  the  stroke.  The  drop  in  pressure  from  1  to  3  was  caused  by 
wire-drawing,  due  to  small  steam  pipes  and  passages  and  to  gradual 
closure  of  the  valve.  The  height  of  3'  above  o'v'  shows  the  back  pres- 
sure at  cut-off,  the  exhaust  still  taking  place  through  the  small  opening 
at  s'. 

The  next  important  position  is  shown  in  Fig.  5,  where  the  port  s' 
closes  and  compression  is  just  about  to  occur,  the  height  of  the  point  4' 
above  o'v'  indicating  the  pressure  at  the  point  of  compression.  Expan- 
sion is  taking  place  on  the  left-hand  side,  the  pressure  falling  to  the 
point  4.  At  mid  position  of  the  valve,  Fig.  6,  expansion  is  taking  place 
on  the  left,  the  pressure  falling,  due  to  increase  of  volume  of  the  steam. 
On  the  right,  compression  is  taking  place,  causing  the  pressure  to  rise 
by  decreasing  the  volume  of  the  entrapped  steam.  In  Fig.  7  is  shown 
the  position  of  the  piston  when  the  point  of  release  occurs  at  s; 
the  steam  is  just  about  to  be  released  after  having  pushed  the  piston  to 
this  point  of  the  stroke,  the  pressure  at  release  being  shown  by  the 
height  of  6  above  ov.  Compression  is  still  taking  place  on  the  other 
side. 

As  the  motion  continues,  exhaust  takes  place  at  s  and  the  pressure 
falls.  The  steam  edge  comes  in  line  with  the  edge  of  the  steam  port  at 
s',  Fig.  8;  steam  is  just  about  to  be  admitted,  the  pressure  having  been 
raised  by  compresion  to  7'. 

When  the  piston  reaches  the  end  of  the  stroke,  Fig.  9,  the  valve  is 
open  to  the  extent  of  the  exhaust  lead  at  s  and  to  the  extent  of  the 
steam  lead  at  s'.  The  pressure  at  8  has  fallen  nearly  to  the  back  pres- 
sure, and  at  8'  has  rapidly  risen  to  the  boiler  pressure  from  the  terminal 
pressure  of  compression  at  7'.  If  the  return  stroke  were  now  made,  the 
other  part  of  the  diagram  shown  in  this  figure  would  be  formed,  thus 
completing  the  diagram  for  each  side  of  the  piston. 

Any  point  on  this  diagram  represents  the  pressure  at  a  certain  posi- 
tion of  the  piston.  The  upper  curve  of  the  diagram  shows  the  varying 
pressure  pushing  the  piston  forward,  and  the  lower  curve  shows  the 
varying  pressure  tending  to  hold  the  piston  back.  The  diagram  from 
one  end  of  the  cylinder  thus  shows  the  pressure  on  one  side  of  the  piston 
during  two  strokes.  In  balancing  engines  it  is  desirable  to  know  the 
effective  pressure  pushing  the  piston  forward  at  each  point  of  the  stroke 
and  this  is  obtained  by  combining  the  top  line  of  one  diagram  with  the 
bottom  line  of  the  other. 


392 


APPENDIX 


a' 


Exhaust  Taking  Place 


p1        Point  of  Admission 


o' 


Fig.  6. 


Steam  Lead 


fig.  9. 


APPENDIX  393 

SOME  USEFUL  NOTES  AND  CONSTANTS. 

Wrought  iron  weighs  480  pounds  per  cubic  foot.     Then,  il|?.  —  36 

cubic  inches  weigh  10  pounds.     Therefore, 

volume  in  cubic  inches       1Q  _  weight  IQ  pounds 

36 
Or, 

sectional  area  in  sq.  inches  x  length  in  feet  x  10  _  weight. 

3 

Whence: 

Sectional  area  in  sq.  inches  —  weight  per  lineal  foot  x  TO  • 
Weight  per  lineal  foot  —  Sectional  area  x  V0  • 

The  weight  of  steel  per  cubic  foot,  490  Ibs.,  is  just  2%  greater  than 
that  for  wrought  iron.  Hence,  a  lineal  foot  of  steel,  one  square  inch  in 
section,  weighs  -y-  x  1-02  —  3.4  Ibs.  We  have,  therefore,  for  steel : 

Sectional  area  in  square  inches  =  weight  per  jineal  foot . 

3.  * 

Weight  per  lineal  foot  =  sectional  area  x  3.4. 

Cast  iron,  which  weighs  450  Ibs.  per  cubic  foot,  is  just  ^  lighter  than 
wrought  iron. 

Wrought  iron  is  9.6  as  heavy  as  white  oak,  10.7  as  heavy  as  yellow 
pine,  and  19.2  as  heavy  as  white  pine. 
1  gallon  =  231  cubic  inches. 
1  gallon  of  water  weighs  8.355  pounds. 
1  pound  avoirdupois  =  7,000  grains  =  453.6  grammes. 
1  cubic  foot  of  water  weighs  62.5  pounds,  for  practical  purposes. 
1  knot  =  6,080  feet. 

1  cubic  foot  of  air,  under  atmospheric  conditions,  weighs  0.08  pound. 
1  horse-power  =  33,000  ft.-lbs.  per  minute  =  746  watts  =  0.746  K.  W. 
1  K.  W.  =  1.34  H.  P. 
Watts  =  volts  x  amperes. 
TT  =  3.1416  =  22/7,  approximately. 

The  base  of  the  Naperian,  or  hyperbolic,  logarithms  is  2.7183. 
To  convert  common  into  Naperian  logarithms,  multiply  by  2.3026. 
V3  =  1.732  and  V2~=  1.414. 
A  column  of  water  2.3  feet  high  corresponds  to  a  pressure  of  one  pound 

per  square  inch. 
A  column  of  mercury  2.04  inches  high  corresponds  to  a  pressure  of  one 

pound  per  square  inch. 

Inches  of  vacuum  x  0-49  =  pounds  pressure. 

1  foot     =  0.305  metre.  1  sq.  inch  =  6.541  sq.  cms. 

1  inch    =25.4  millimetres.  1  pound     =  0.4536  kilogrammes. 

1  sq.  ft.  =  0.0929  sq.  metre.  1  kilo.        =2.2  pounds. 

1  mile        =1609.3  metres  =  1.61  kilometres. 

1  cu.  inch  =  16.387  cu.  cms. 

1  cu.  foot  =0.0283  cu.  metre. 


INDEX 

PAGE 

Absolute  temperature    4 

unit  of  force 383 

zero  of  temperature 4 

Accumulated  work   386 

Adiabatic  expansion    98 

Admission  line  of  indicator  diagram 80 

Air,  quantity  necessary  for  combustion 24 

Alloys    246 

American  Society  of  Mechanical  Engineers,  boiler  trial  code  of 164 

Angular  advance  of  the  eccentric 53 

Angularity  of  connecting-rod 66 

Anthracite  coal  22 

Appendix 383 

Area  of  test  pieces,  reduction  in 256 

Atmospheric  line  of  indicator  diagram 80 

Automatic  cut-off   60 

Axis,  neutral   312 

Back  pressure  line  of  indicator  diagram 81 

Barrel  calorimeter  106 

Beam,  cantilever    282 

resting  on  three  supports 333 

simple 282 

Beams,  continuous 333 

deflection  of   329-335 

resilience  of    338-340 

standard  I  317 

table  of  relative  strength  of 335 

theory  of   311 

theory  of,  important  assumptions  made 314 

with  fixed  ends 331 

Belt  and  pulley,  f rictional  resistance  between 208 

Belt,  thickness  of 207 

Belting   205 

Belting,   strength   of 210 

Belts,  centrifugal  action  in 211 

creeping  of  210 

transmission  of  power  by 210 

Bending  moments 282 

Bending  moments,  general  case  of 282 

Bending-moment  diagrams   284 

Bending-moment  diagram,  the  funicular  polygon  a 352 

Bending-moment  and  shear  diagrams,   for   different  conditions   of 
loading  . ,  296-302 


396  INDEX 

PAGE 

Bending-moment  and  shear  diagrams  of  cantilevers 293 

Bending-moment  and  shear  diagrams,  relations  between 288 

Bituminous  coal  22 

Boiler,  code  for  conducting  trials  of 30 

design  181 

efficiency 40 

girder  stays  for 185 

grate  surface  of 183 

heating  surface  of 183 

horse-power   of    30 

sectional  area  of  tubes  and  of  area  over  bridge  wall 183 

steam  space  in 130 

strength  of,  longitudinally  and  circumferentially 183 

test,  record  of 175 

test,  report  of 176 

thermal  efficiency  of  engine  and 39 

The  Roberts 186-189 

trials,  rules  for  conducting,  code  of  1899 164 

volume  of  steam  space  and  of  combustion  chamber 183 

Boilers,  classification  of    179 

efficiency  of  engine  and 38 

locomotive,  marine,  stationary,  sectional 179 

relative  advantages  of  different  types 180 

Scotch    180 

steam  179 

Boiling  point  : 13 

Bow's  system  of  lettering 345 

Boyle's  law  3 

Brake  horse-power   10 

Brake,  Prony 161 

Brazing    231 

Brazing  of  cast-iron 236 

Calorimeter,  barrel    106 

separating    109 

tests    106-109 

throttling  107 

Calorimetry  8 

Cantilever  beam  282 

Cantilevers,  bend  ing-moment  and  shear  diagrams  of 293 

Carbon,  graphitic   232 

in  fuel 23 

on  cast  iron,  influence  of 233 

Carnot,  perfect  heat  engine  of 33, 110 

Castings,  inspection  of 236 

Cast  iron 232 

Cast  iron,  brazing  of 236 


INDEX  397 

PAGE 

Cast  iron,  influence  of  carbon  on 233 

influence  of  chromium  on 235 

influence  of  manganese  on  235 

influence  of  silicon  on 233 

influence  of  sulphur  on  234 

malleable  236 

shrinkage   of    235 

strength  and  hardness  of 235 

uses   of   in  engineering 236 

Center,  placing  engine  on 55 

of  gravity 268 

of  gyration 306 

of  pressure 251 

Centrifugal  action  in  belts 211 

Centrifugal  force    385 

Centripetal  force 385 

Charles'   law    3 

Checks  as  to  the  accuracy  of  shear  diagrams 302 

Chromium,  on  cast  iron,  influence  of 235 

Circles,  pitch  217 

Clearance    82 

Clearance  line  of  indicator  diagram 81 

Clearance,  on  ratio  of  expansion,  influence  d¥ 83 

Coal    22 

Coal,  anthracite 22 

bituminous   22 

pounds  per  I.  H.  P.  per  hour 38 

Code,  short  of  1899,  for  conducting  boiler  trials 164 

Coefficient  of  elasticity 311 

Coke    22 

Columns    320 

Columns,  the  design  of 322 

Combination  of  the  laws  of  Boyle  and  Charles 5 

Combining  diagrams  of  stage  expansion  engines 155 

Combustion    23 

Combustion,  efficiency  of  40 

quantity  of  air  necessary  for 24 

Compounding  38 ' 

Compression    47 

Compression  curve 81 

Compression  tests 264 

Concrete    247 

Concrete,  adhesion  of  to  steel 248 

steel   247 

Condensation  and  production  of  vacuum 116 

Condensation  and  re-evaporation  of  steam  in  cylinders 35 

Condenser,  jet,  surface 116 

Condensing  water,  weight  required  per  pound  of  steam 117 


398  INDEX 


PAGE 

Connecting-rod,  action  of  crank  and 64 

angularity  of    66 

Conservation  of  energy 9 

Constants,  some  useful  notes  and 393 

Continuous  beams    333 

Continuous  expansion  type  of  stage  expansion  engines 71 

Convection,  transfer  of  heat  by 11 

Conversion  of  motion 64 

Copper    231 

Corrugated  furnace  184 

Crank  and  connecting-rod,  action  of 64 

Crank  and  piston,  relative  positions  of 67 

Crank-pin-piston  velocity  diagram 133 

Creeping  of  belts 210 

Curve,  compression 81 

Curves  of  adiabatic,  saturated,  and  isothermal  expansion,  construc- 
tion of  99- 

Cut-off  47 

Cut-off,  automatic 60 

Cylinder  ratios 125 

Cylinder,  size  of  for  given  power 124 

Cylinders,  liquefaction  in   113 

jt 

Dangerous  section    314 

Deficient  frame,  a 377 

Deflection  of  beams 329-335 

De  Laval  turbine 191-198 

Diagram,  force    345 

ideal  indicator 79 

indicator,  horse-power  from 92 

indicator,  in  preliminary  design 121 

indicator,  measure  of  work  in  cylinder 94 

reciprocal   363 

reciprocal,  the  drawing  of 365 

stress-strain 259 

Diagrams,  bending-moment    284 

bending-moment  and  shear,  of  cantilevers 293 

bending-moment  and  shear,  relations  between 287 

indicator,  interpretation  of 79 

indicator,  steam  per  I.  H.  P.  per  hour  from 150 

of  stage  expansion  engines,  combining 155 

Divergent  nozzle  of  De  Laval  turbine 193 

Dryness  fraction  of  steam 105 

Eccentric,  angular  advance  of 53 

Eccentric  arm    49 

Eccentric  radius    49 

Eccentric  rods,  crossed 57 

Eccentric  rods,  open   57 


INDEX  399 

PAGE 

Eccentric,  shifting   60 

the   49 

throw  of   ' 

virtual    58 

Economy  of  the  steam  engine  and  boiler 150 

Efficiencies  of  engines  and  boilers 38, 103 

Efficiency  ' 

Efficiency,  boiler    40> 103 

engine 11G 

losses  affecting  34 

maximum   34 

Efficiency  of  combustion    ' 

Efficiency  of  engine,  mechanical  40 

Efficiency  of  heating  surface 40 

Efficiency,  relative  engine  40 

thermal 39 

total  of  system 4i 

Elastic  limit    256 

Elasticity,  coefficient  of   3 

modulus  of   Ml 

Elongation  in  test  pieces 256 

Energy   - 7 

Energy,  conservation  of  • 9 

kinetic    7 

potential    7 

Engine  and  boiler,  thermal  efficiency  of 39 

Engine,  dead  points  of 55 

Engine  efficiency  110 

Engine  efficiency,  relative  40 

Engine,  mechanical  advantage  of  stage  expansion 131 

mechanical  efficiency  of  40 

placing  on  center 55 

reciprocating,  compared  with  steam  turbine 202 

thermal  efficiency  of 39 

Engines  and  boilers,  efficiencies  of 38 

Evaporation,  factor  of 30 

unit  of  30 

Evaporative  effect  of  mineral  oil 23 

Evaporative  power  of  fuel 26 

Exhaust  line  of  indicator  diagram 81 

Expansion,  adiabatic,  saturated  steam,  isothermal,  hyperbolic 98 

engines,  relative  merits  of  the  simple  and  the  stage. ...    70 

engines,  stage   70 

influence  of  clearance  on  the  ratio  of 83 

ratio  of  83 

ratio  of  in  stage  expansion  engines 85 


400  INDEX 


PAGE 

Factor  of  evaporation  30 

Factor  of  safety 256 

Factors,  mean  pressure » 125 

Feed  water,  weighing  the 175 

Fink  truss,  the 376 

Fixed  ends,  beams  with 331 

Foot  pound 10 

Force  , 9 

Force,  absolute  unit  of 383 

centrifugal   385 

centripetal   385 

diagram 345 

gravitation  unit  of    383 

measurement  of 383 

Frame,  a    362 

a  deficient 377 

a  redundant 377 

Framed  structures   362 

Fuel,  evaporative  power  of 26 

Fuels   22 

Fuels,  total  heat  of  combustion  of 26 

Funicular  polygon   348 

Funicular  polygon,  a  bending-moment  diagram 352 

Furnace,  Adamson  ring  and  Bowling  hoop  for 184 

corrugated 184 

Gas,  a  3 

Gas  expanding  within  a  cylinder,  mean  pressure  of 96 

Gases,  kinetic  theory  of 3 

Gay  Lussac,  law  of 3 

Gear,  of  lathe,  back  225 

wheels,  horse-power  transmitted  by 222 

wheels,  transmission  of  power  by 220 

Gearing,  lathe,  compound   225 

lathe,  simple   223 

Generation  of  steam  under  constant  pressure 14 

Generation  of  steam  under  constant  volume 17 

Girder,  N,  or  Linville 375 

Girder  stays  for  boilers 185 

Girder,  Warren  373 

Governor,  shaft 61 

Graphic  statics 345 

Graphitic  carbon    232 

Grate  surface  183 

Gravitation  unit  of  force 383 

Gravity,  center  of 268 

Gyration,  radius  of 306 


INDEX  401 

PAGE 

Harmonic  motion,  simple 65 

Heat,  application  of  to  water 11 

kinetic  theory  of 6 

latent    13 

loss  of  by  radiation 11 

material  theory  of 6 

mechanical  equivalent  of 0 

necessary  to  evaporate  a  pound  of  water 29 

of  combustion  of  fuels,  total 26 

of  steam,  total 16 

radiant 11 

rejected  into  the  condenser 117 

sensible    12 

specific  8 

transfer  by  convection 11 

Heat  engine,  Carnot's  perfect 33 

Heating  surface    183 

Heating  surface,  efficiency  of 40 

Horse-power    10 

Horse-power,  boiler    30 

brake    10 

from  indicator  diagrams 92 

indicated    92 

transmitted  by  gear  wheels 222 

transmitted  by  shafts 325 

Hyperbolic  expansion   98 

I  beams,  standard 317 

Ideal  engine  Ill 

Ideal  engine,  pounds  of  steam  per  I.  H.  P.  per  hour,  table  of 112 

Ideal  indicator  diagram  79 

Indicator  diagram   75 

formation  of  386 

horse-power  from   92 

ideal    79 

in  preliminary  design 121 

measure  of  work  performed  in  the  cylinder 94 

Indicator  diagrams,  interpretation  of  79 

Indicator  diagrams,  steam  per  I.  H.  P.  per  hour  from 150 

Indicator,  location  of 80 

Indicator  springs    76 

Indicator,  Tabor    77 

the 75 

Indicated  horse-power 92 

Indicated  horse-power,  pounds  of  coal  per  hour  per 38 

pounds  of  steam  per  hour  per 39, 123 

thermal  units  per  minute  per 39 

Inequality  in  the  motion  of  the  piston 66 

26 


402  INDEX 


PAGE 

Inertia,  moment  of  304 

polar  moment  of 305 

Inflection,  points  of 330 

Iron,  cast  232 

wrought  237 

test  pieces  for 256 

Isothermal  expansion   98 

Jacketing,  steam  37 

Joy,  valve  gear  of 63 

Kinetic  theory  of  gases 3 

Kinetic  theory  of  heat .- . . .      € 

Lap,  definition  of 52 

Latent  heat   13 

Latent  heat  of  steam 16 

Lathe,  back  gear  of 225 

gearing,  compound  225 

gearing,  simple 223 

Lead    245 

Lead  of  valve    46 

Lead  of  valve,  definition  of 52 

Lettering,  Bow's  system  of 345 

Limit,  elastic  256 

Line,     atmospheric,     admission,     steam,     exhaust,     back     pressure, 

vacuum,   clearance    80,    81 

Link  motion 56 

Link,  Stephenson's 59 

Linville,  or  N,  girder 375 

Liquefaction  in  cylinders 113 

Machine,  mechanical  advantage  of 219 

mechanical  efficiency  of  219 

Malleable  cast  iron 236 

Manganese,  influence  of  on  cast  iron   235 

influence  of  on  steel    241 

Marshall,  valve  gear  of 63 

Mass 383 

Materials   231 

Materials,  different  kinds  of  tests  of 255 

testing    255 

Mean  effective  pressure  from  indicator  diagrams 90 

Mean  pressure  factors    125 

Mean  pressure  in  stage  expansion  engines 74 

Mean  pressure  of  gas  expanding  within  a  cylinder 96 

Mean  pressure  with  adiabatic,  saturated  steam,  and  hyperbolic  ex- 
pansion      98 

Measurement  of  force . .  . .  383 


INDEX  403 

PAGE 

Mechanical  advantage  of  a  machine  219 

Mechanical  efficiency  of  an  engine 40 

Mechanical  efficiency  of  a  machine   219 

Mechanical  equivalent  of  heat 9 

Mild  steel,  tension  test  of 261 

Mild,  or  structural,  steel 239 

Mineral  oil 23 

Mineral  oil,  evaporative  effect  of 23 

Modulus  of  elasticity    311 

Modulus  of  resilience 337 

Modulus,  section  for  bending 314 

section  for  torsion 324 

Moment  of  inertia 304 

Moment,  maximum  equivalent  twisting 324 

resisting    313 

Moments   267 

Moments,  bending  282 

bending,  general  case  of 282 

Motion,  conversion  of 64 

link    56 

simple  harmonic  65 

N,  or  Linville,  girder 375 

Neutral  axis    312 

Neutral  surface    207,  312 

Notes  and  constants,  some  useful 393 

Nozzle  of  De  Laval  turbine,  divergent 193 

Oil,  mineral   23 

Oil,  mineral,  evaporative  effect  of 23 

Phosphorus,  influence  on  steel 241 

Piston,  inequality  in  motion  of 66 

relative  position  of  the  crank  and 67 

Piston  speed     125 

Piston  valve 62 

Pitch  circles  217 

Pitch  of  teeth 217 

Plasticity 259 

Point,  boiling 13 

Point  of  cut-off,  release,  exhaust  closure 81 

Points  of  inflection 330 

Polar  moment  of  inertia 305 

Polygon,  funicular    343 

funicular,  a  bending-moment  diagram 352 

Port,  steam,  dimensions  of 130 

Power jo 

Power,  horse    10 

horse,  indicated 92 

horse,  transmitted  by  gear  wheels .  222 


404  INDEX 


PAGE 

Power,  size  of  cylinder  for  a  given 124 

transmission  of  by  belts 210 

transmission  of  by  gear  wheels 220 

Pressure,  formula  connecting  temperature  with 18 

formula  connecting  volume  with 19 

line,  back 81 

mean  effective  from  indicator  diagram 90 

mean  factors  of  125 

mean  with  adiabatic  expansion,  saturated  steam  expan- 
sion, hyperbolic  expansion 98 

Prony  brake 161 

Pulley,  frictional  resistance  between  a  belt  and 208 

Quick  running :    38 

Radial  valve  gears 63 

Radiation,  loss  of  heat  by 11 

Radius  of  eccentric 49 

Radius  of  gyration 306 

Ratio  of  expansion  83 

Ratio  of  expansion,  influence  of  clearance  on 83 

Ratio  of  expansion  in  stage  expansion  engines 85 

Ratio,  velocity  219 

Ratios,  cylinder 125 

Receiver  type  of  stage  expansion  engines 71 

Reciprocal  diagram   363 

Reciprocal  diagram,  drawing  the 365 

Record  of  a  boiler  test 176 

Reduction  in  area  of  test  pieces 256 

Redundant  frame,  a 377 

Relative  strength  of  beams,  table  of 335 

Release  48 

Report  of  a  boiler  test 176 

Representative  strengths  and  weights  of  materials,  table  of 263 

Resilience    337 

Resilience  from  sudden  loads  and  shocks 337 

Resilience,  modulus  of 337 

Resilience  of  beams    338-340 

Resilience  of  shafts  under  torsion 340 

Resisting  moment • 313 

Ritter  section  method  of  determining  stresses 370 

Roberts  boiler,  the 186-189 

Running,  quick   38 

Rules  for  conducting  boiler  trials,  code  of  1899 164 

Safety,   factor  of 256 

Saturated  steam •  • ^6 

Saturated  steam  expansion • 98 

Section,  dangerous     • 314 


INDEX  405 

PAGE 

Section  modulus  for  bending  314 

Section  modulus  for  torsion  , 324 

Separating  calorimeter  109 

Setting  slide  valves 55 

Shaft  governor   61 

Shafts    324 

Shafts,  horse-power  transmitted  by 325 

maximum  equivalent  twisting  moment  of 324 

under  torsion,  resilience  of 340 

Shear    285 

Shear   and   bending-moment   diagrams   for   different   conditions   of 

loading 296-302 

Shear  diagrams 285 

Shear  diagrams,  checks  to  the  accuracy  of 302 

relations  between  bending-moment  and 288 

Shear,  first  a>derivative  of  the  moment  is  the 315 

general  case  of 285 

Shifting  eccentric 60 

Shrinkage  of  cast  iron 235 

Silicon,  influence  of  on  cast  iron 233 

influence  of  on  steel 241 

Slide  valve,  steam  distribution  by. 386 

Smoke,  cause  and  prevention  of 24 

Specific  heat    8 

Speed,  piston   125 

Stage  expansion  engines 70 

Stage  expansion  engines,  combining  diagrams  of 155 

continuous  expansion  type 71 

mean  pressure  in 74 

mechanical  advantage  of 131 

ratio  of  expansion  in 85 

receiver  type  of ;    71 

relative  merits  of  the  simple  and 70 

Statics,  graphic 345 

Steam,  distribution  of  by  slide  valve 386 

dryness  fraction  of 105 

generation,  of  under  constant  pressure 14 

generation,  of  under  constant  volume 17 

latent  heat  of 16 

per  I.  H.  P.  per  hour  from  actual  indicator  diagrams 150 

per  I.  H.  P.  per  hour  for  ideal  engine,  table  of 112 

pounds  per  I.  H.  P.  per  hour 39, 112 

superheated    36 

total  heat  of 16 

weight  of  condensing  water  required  per  pound  of 117 

Steam  boilers    179 

Steam  engine  and  boiler,  economy  of  the 150 

Steam  jacketing    .    37 


406  INDEX 

PAGE 

Steam  line  of  indicator  diagram 80 

Steam  port,  dimensions  of 130 

Steam  space  in  boiler  130 

Steam  turbine,  the 190 

Steel   238 

Steel,  Bessemer  process 239 

chrome  244 

crucible 239 

flange,  or  rivet 242 

for  castings   243 

for  shafting   243 

influence  of  manganese  on   241 

influence  of  phosphorus  on  241 

influence  of  silicon  on   241 

mechanical  properties  of 242 

mild,  or  structural 239 

nickel    244 

open-hearth  process  239 

proportions  of  carbon  in 238 

semi    242 

shell    243 

special  properties  of  243 

table  of  temperatures  and  colors  corresponding  to  tempers  of.  244 

tempering  of   243 

test  pieces  for   257,  258 

tungsten,  or  mushet   244 

in  marine  construction,  uses  of 245 

Steel-concrete 247 

Steel-concrete  and  steel  and  concrete  construction,  distinction  be- 
tween  249 

Steel-concrete  beam,  design  of 250 

Steels,  open-hearth  and  Bessemer  compared 241 

special    244 

Stephenson  link  59 

Strain    255-311 

Strength,  ultimate  256 

Stress    J11 

Stress,  rule  for  determining  kind  of 368 

Stress-strain  diagram   259 

Stroke   125 

Stroke,  forward,  outward,  top,  head,  down,  return,  inward,  bottom, 

crank,  up 143 

Structure,  a  framed  362 

equilibrium  of  a  framed 362 

load  on  a  framed 362 

reaction  on  a  framed 362 

Sulphur,  influence  of  on  cast  iron    , 234 

influence  of  on  steel    241 


INDEX  407 

PAGE 

Superheated  steam 36 

Surface;  grate    183 

heating    183 

neutral   207,  312 

System  of  lettering,  Bow's 345 

System,  total  efficiency  of 41 

Table  of  relative  strengths  of  beams 335 

Table  of  representative  strengths  and  weights  of  materials 263 

Table  of  temperatures  and  colors  corresponding  to  tempers  in  steel .  244 

Tabor    indicator 77 

Temper  in  steel,  table  of  temperatures  and  colors  corresponding  to. . .  244 

Temperature 7 

Temperature,  absolute 4 

conversion  of  from  one  scale  to  another 8 

formula  connecting  pressure  with 18 

Tempering    243 

Test  of  mild  steel,  tension 261 

Test  piece  for  iron     256 

Test  pieces  for  steel  and  other  materials 257,  258 

Tests,  calorimeter    106-109 

compression    264 

engine  and  boiler 162 

Theory  of  beams,  the    311 

important  assumptions  made 314 

Thermal  efficiency    39 

Thermal  efficiency  of  engine 39 

Thermal  efficiency  of  engine  and   boiler 39 

Thermal   unit    8 

Thermodynamics   8 

Thermodynamics,  first  law  of 8 

second   law  of 9 

Throttling  calorimeter 107 

Throw  of  eccentric   49 

Timber    246 

Travel  of  valve 49 

Ultimate  strength    256 

Unit,  British  thermal 8 

of  evaporation    30 

of  work   10 

Units  per  I.  H.  P.  per  minute,  thermal 39 

Vacuum,  condensation  and  production  of 116 

Vacuum  line  of  indicator  diagram 81 

Value  of  a  train  of  wheels 219 

Valve  and  its  motion,  the 45 

Valve,  lap  of  the 52 

lead  of  the  .    52 


408  INDEX 

PAGE 

Valve,  piston   62 

slide,  distribution  of  steam  by 386 

slide,  setting  of  55 

travel  of  the 49 

Valve  gear,  Joy's    63 

Marshall's    63 

Valve  gears,  radial 63 

Velocity  ratio 219 

Virtual  eccentric 58 

Vis  viva    386 

Volume,  formula  connecting  pressure  and 19 

Warren  girder 373 

Water,  heat  necessary  to  evaporate  a  pound  of 29 

Water,  weighing  the  feed 175 

Weight  of  a  body,  the 383 

Westinghouse-Parsons  turbine,  the  198-202 

Wheels  in  train 217 

Wheels,  transmission  of  power  by  gear 220 

Wire  drawing  88 

Wood 23 

Work    9 

Work,  accumulated    386 

percentage  of  applied  to  shaft 41 

unit  of 10 

Wrought  iron    237 

Wrought  iron,  proportion  of  carbon  in 237 

special  properties 238 

uses  in  marine  engineering 238 

Zero  of  temperature,  absolute 4 

Zeuner  valve  diagram 137 

Zinc    .  245 


UNIVERSITY  OF  CALIFORNIA  LIBRARY 
BERKELEY 

Return  to  desk  from  which  borrowed. 
This  book  is  DUE  on  the  last  date  stamped  below. 


APR 


LD  21-100m-9,'47(A5702sl6)476 


